Recent questions in Exponential growth and decay

Algebra IAnswered question

Keenan Conway 2022-09-24

rate of change proportional to the diffrence between itself and a fixed value $\frac{dy}{dx}=k(y-a)$

So I have a question regarding exponential growth/decay problem formulated in my book.

In my book they say:

"Sometimes an exponential growth or decay problem will involve a quantity that changes at a rate proportional to the difference between itself and a fixed point:

$\frac{dy}{dx}=k(y-a)$

In this case, the change of dependent variable u(t)=y(t)−a should be used to convert the differential equation to the standard form. Observe that u(t) changes at the same rate as $y(t)(ie,\frac{du}{dy}=\frac{dy}{dt}) so it satisfies

$\frac{du}{dt}=ku$

I am having a hard time grasping what they mean. And I would hope for somebody to expand on this notion.

More specifically how does $\frac{dy}{dx}=k(y-a)$ relate to $u(t)=y(t)-a$?

So I have a question regarding exponential growth/decay problem formulated in my book.

In my book they say:

"Sometimes an exponential growth or decay problem will involve a quantity that changes at a rate proportional to the difference between itself and a fixed point:

$\frac{dy}{dx}=k(y-a)$

In this case, the change of dependent variable u(t)=y(t)−a should be used to convert the differential equation to the standard form. Observe that u(t) changes at the same rate as $y(t)(ie,\frac{du}{dy}=\frac{dy}{dt}) so it satisfies

$\frac{du}{dt}=ku$

I am having a hard time grasping what they mean. And I would hope for somebody to expand on this notion.

More specifically how does $\frac{dy}{dx}=k(y-a)$ relate to $u(t)=y(t)-a$?

Algebra IAnswered question

Julia Chang 2022-09-23

Why below question is considered as Linear and not exponential growth?

A ball falls from a height of 2 meters onto a firm surface and jumps after each impact each back to 80% of the height from which it fell. ¨ Set up the function, which indicates the height of the ball after the nth impact reached. How high does the ball jump after the 5th impact?

Below question is considered as a linear growth and not exponential growth . I dont understand why its linear growth or decay.

$$y=2\ast {0.8}^{5}$$

For the percentage neither it add to 1 nor minus from 1.

I wish to ask why it's not $y=2\ast {.2}^{5}$ (I took 80% as decay and did 1−.8=.2)

A ball falls from a height of 2 meters onto a firm surface and jumps after each impact each back to 80% of the height from which it fell. ¨ Set up the function, which indicates the height of the ball after the nth impact reached. How high does the ball jump after the 5th impact?

Below question is considered as a linear growth and not exponential growth . I dont understand why its linear growth or decay.

$$y=2\ast {0.8}^{5}$$

For the percentage neither it add to 1 nor minus from 1.

I wish to ask why it's not $y=2\ast {.2}^{5}$ (I took 80% as decay and did 1−.8=.2)

Algebra IAnswered question

sombereki51 2022-09-22

A colleague came across this terminology question.

What are the definitions of exponential growth and exponential decay? In particular:

1) Is $f(x)=-{e}^{x}$ exponential growth, decay, or neither?

2) Is $g(x)=-{e}^{-x}$ exponential growth, decay, or neither?

Consider $f(x)=A{e}^{kx}.$. I can't find any sources that specify A>0. My answer is that:

$f$ exhibits

1. exponential growth for A>0,k>0, and

2. exponential decay for A>0,k<0

whereas $|f|$ exhibits

3. exponential growth for A<0,k>0, and

4. exponential decay for A<0,k<0.

In case (3) we shouldn't call $f$ an exponential growth function without noting that it is "negative growth". Also it wouldn't be called it an exponential decay function without specifying the "direction of decay", so it is neither.

In case (4) it's neither as well. One should specify that it is the magnitude of $f$ which decays exponentially although $f$ is increasing in value. Although $f$ is increasing in value, is it growing? It seems odd to say it is exponentially growing.

It just doesn't sit right with me to refer to a function as growing if it is decreasing in value. Certainly, it's magnitude may be growing.

Next consider a function with exponential asymptotic behavior (e.g. logistic) so that as $x\to \mathrm{\infty},$ $f(x)\approx A{e}^{-kx}+C$ for some k>0. I feel the best way to describe this would be "exponential decay towards C" with a qualification as being from as being from above or below depending on the sign of A.

If someone is to just use the terminology "exponential growth (decay)", it implies $f(x)=A{e}^{kx}$ with positive A and k>0 (k<0) unless there is a specific context or further clarification as to what the actual nature of the function is.

What are the definitions of exponential growth and exponential decay? In particular:

1) Is $f(x)=-{e}^{x}$ exponential growth, decay, or neither?

2) Is $g(x)=-{e}^{-x}$ exponential growth, decay, or neither?

Consider $f(x)=A{e}^{kx}.$. I can't find any sources that specify A>0. My answer is that:

$f$ exhibits

1. exponential growth for A>0,k>0, and

2. exponential decay for A>0,k<0

whereas $|f|$ exhibits

3. exponential growth for A<0,k>0, and

4. exponential decay for A<0,k<0.

In case (3) we shouldn't call $f$ an exponential growth function without noting that it is "negative growth". Also it wouldn't be called it an exponential decay function without specifying the "direction of decay", so it is neither.

In case (4) it's neither as well. One should specify that it is the magnitude of $f$ which decays exponentially although $f$ is increasing in value. Although $f$ is increasing in value, is it growing? It seems odd to say it is exponentially growing.

It just doesn't sit right with me to refer to a function as growing if it is decreasing in value. Certainly, it's magnitude may be growing.

Next consider a function with exponential asymptotic behavior (e.g. logistic) so that as $x\to \mathrm{\infty},$ $f(x)\approx A{e}^{-kx}+C$ for some k>0. I feel the best way to describe this would be "exponential decay towards C" with a qualification as being from as being from above or below depending on the sign of A.

If someone is to just use the terminology "exponential growth (decay)", it implies $f(x)=A{e}^{kx}$ with positive A and k>0 (k<0) unless there is a specific context or further clarification as to what the actual nature of the function is.

Algebra IAnswered question

demitereur 2022-09-22

I'm trying to graph a simple response function: 1/(1-0.5s^-1)

Now, I know that the function can also be written as: s/(s-0.5)

So I tried plotting the step and impulse responses in Matlab:

sys = tf([1 0],[1 -0.5])

figure(1);

step(sys);

figure(2);

impulse(sys);

However, both graphs look the same (can't post images of my graphs, I need more rep to do it).

Both graphs have exponential growth, but shouldn't the impulse response look like exponential decay?

Now, I know that the function can also be written as: s/(s-0.5)

So I tried plotting the step and impulse responses in Matlab:

sys = tf([1 0],[1 -0.5])

figure(1);

step(sys);

figure(2);

impulse(sys);

However, both graphs look the same (can't post images of my graphs, I need more rep to do it).

Both graphs have exponential growth, but shouldn't the impulse response look like exponential decay?

Algebra IAnswered question

Camila Brandt 2022-09-22

Differential equation: the law of natural growth and the law of natural decay

I understand that $\frac{dy}{dx}=k\ast y$ and when k>0 this is the law of natural growth and when k<0 this is the law of natural decay, but my textbook gives an example of radioactive decay as follows which confuses me:

Radioactive substances decay by spontaneously emitting radiation. If is the mass remaining from an initial mass of the substance after time t, then the relative decay rate

$\frac{-1}{m}\frac{dm}{dt}$

has been found experimentally to be constant. (Since $\frac{dm}{dt}$ is negative, the relative decay rate is positive.) It follows that:

$\frac{dm}{dt}=km$

where k is a negative constant. In other words, radioactive substances decay at a rate proportional to the remaining mass. This means that we can use to show that the mass decays exponentially:

$m(t)=m(0){e}^{kt}$ (This equation I understand and accept, the above two confuses me)

Now eventually when the example becomes numerical, then of course k becomes a negative number since it is natural decay. However, the above general notation explanation confuses me because it keeps changing the sign of k=relative decay rate(from negative(1) to positive(2), but eventually when numerically worked it turns out to be a negative constant since it's natural decay). I know that k must be less than zero for decay but I'm just trying to fully grasp the notation signs that the textbook uses in the explanation above.

I understand that $\frac{dy}{dx}=k\ast y$ and when k>0 this is the law of natural growth and when k<0 this is the law of natural decay, but my textbook gives an example of radioactive decay as follows which confuses me:

Radioactive substances decay by spontaneously emitting radiation. If is the mass remaining from an initial mass of the substance after time t, then the relative decay rate

$\frac{-1}{m}\frac{dm}{dt}$

has been found experimentally to be constant. (Since $\frac{dm}{dt}$ is negative, the relative decay rate is positive.) It follows that:

$\frac{dm}{dt}=km$

where k is a negative constant. In other words, radioactive substances decay at a rate proportional to the remaining mass. This means that we can use to show that the mass decays exponentially:

$m(t)=m(0){e}^{kt}$ (This equation I understand and accept, the above two confuses me)

Now eventually when the example becomes numerical, then of course k becomes a negative number since it is natural decay. However, the above general notation explanation confuses me because it keeps changing the sign of k=relative decay rate(from negative(1) to positive(2), but eventually when numerically worked it turns out to be a negative constant since it's natural decay). I know that k must be less than zero for decay but I'm just trying to fully grasp the notation signs that the textbook uses in the explanation above.

Algebra IAnswered question

Logan Knox 2022-09-22

How to solve a Non-algebraic equation?

Whilst working with exponential boom and decay, I encountered a hassle wherein I need to solve an equation related to logarithm. I could not separate or could not make it explicit.

$y=\frac{\mathrm{ln}((1+r)(1-\delta ))}{\mathrm{ln}{\textstyle (}{\textstyle (}1+\frac{r}{y}{\textstyle )}(1-\delta ){\textstyle )}}$. where r and δ are positive, both are smaller than 1.

I need the solution of this in the form of; $y=f(r,\delta )$. I tried numerical solution also but could not find a best way. Any hint or help is appreciated.

Whilst working with exponential boom and decay, I encountered a hassle wherein I need to solve an equation related to logarithm. I could not separate or could not make it explicit.

$y=\frac{\mathrm{ln}((1+r)(1-\delta ))}{\mathrm{ln}{\textstyle (}{\textstyle (}1+\frac{r}{y}{\textstyle )}(1-\delta ){\textstyle )}}$. where r and δ are positive, both are smaller than 1.

I need the solution of this in the form of; $y=f(r,\delta )$. I tried numerical solution also but could not find a best way. Any hint or help is appreciated.

Algebra IAnswered question

saucletbh 2022-09-20

For which α does the integral $$\underset{2}{\overset{+\mathrm{\infty}}{\int}}\frac{{e}^{\alpha x}}{(x-1{)}^{\alpha}\mathrm{ln}x}\mathrm{d}x$$ converge?

For which values of $\alpha $ does the integral

$$\underset{2}{\overset{+\mathrm{\infty}}{\int}}\frac{{e}^{\alpha x}}{(x-1{)}^{\alpha}\mathrm{ln}x}\mathrm{d}x$$

converge?

I'm lost here.

For simplicity, let us denote the above integral as ${I}_{1}$

I was able to prove (I noticed that it's kind of standard exercise though) that:

$${I}_{2}=\underset{2}{\overset{+\mathrm{\infty}}{\int}}\frac{1}{(x-1{)}^{\alpha}{\mathrm{ln}}^{\beta}x}$$

I2 converges for $\alpha >1$ and $\mathrm{\forall}\text{}\beta $.

converges for $\alpha =1$ and $\beta >1$. For $\beta \le 1$ it diverges.

diverges for $\alpha <1$ and $\mathrm{\forall}\text{}\beta $.

So my strategy was to use ${I}_{2}$ to prove the converges/divergence of ${I}_{1}$, by means of inequalities, but I get nothing from this.

For instance I did the following:

For $\alpha >1$ we have that

$$0<\underset{2}{\overset{+\mathrm{\infty}}{\int}}\frac{\mathrm{d}x}{{x}^{\alpha}\mathrm{ln}x}\le \underset{2}{\overset{+\mathrm{\infty}}{\int}}\frac{{e}^{\alpha x}}{(x-1{)}^{\alpha}\mathrm{ln}x}\mathrm{d}x$$

That is, for $\beta =1$

$$0<{I}_{2}\le {I}_{1}$$

From 1. we know that ${I}_{2}$ converges, but this doesn't tell us anything about the convergence of ${I}_{1}$.

The answer given by my textbook is that ${I}_{1}$ converges for $\alpha <0$, but I don't know how to arrive at that conclusion.

For which values of $\alpha $ does the integral

$$\underset{2}{\overset{+\mathrm{\infty}}{\int}}\frac{{e}^{\alpha x}}{(x-1{)}^{\alpha}\mathrm{ln}x}\mathrm{d}x$$

converge?

I'm lost here.

For simplicity, let us denote the above integral as ${I}_{1}$

I was able to prove (I noticed that it's kind of standard exercise though) that:

$${I}_{2}=\underset{2}{\overset{+\mathrm{\infty}}{\int}}\frac{1}{(x-1{)}^{\alpha}{\mathrm{ln}}^{\beta}x}$$

I2 converges for $\alpha >1$ and $\mathrm{\forall}\text{}\beta $.

converges for $\alpha =1$ and $\beta >1$. For $\beta \le 1$ it diverges.

diverges for $\alpha <1$ and $\mathrm{\forall}\text{}\beta $.

So my strategy was to use ${I}_{2}$ to prove the converges/divergence of ${I}_{1}$, by means of inequalities, but I get nothing from this.

For instance I did the following:

For $\alpha >1$ we have that

$$0<\underset{2}{\overset{+\mathrm{\infty}}{\int}}\frac{\mathrm{d}x}{{x}^{\alpha}\mathrm{ln}x}\le \underset{2}{\overset{+\mathrm{\infty}}{\int}}\frac{{e}^{\alpha x}}{(x-1{)}^{\alpha}\mathrm{ln}x}\mathrm{d}x$$

That is, for $\beta =1$

$$0<{I}_{2}\le {I}_{1}$$

From 1. we know that ${I}_{2}$ converges, but this doesn't tell us anything about the convergence of ${I}_{1}$.

The answer given by my textbook is that ${I}_{1}$ converges for $\alpha <0$, but I don't know how to arrive at that conclusion.

Algebra IAnswered question

Kelton Molina 2022-09-20

What are the practical uses of e?

How can e be used for practical mathematics? This is for a presentation on (among other numbers) e, aimed at people between the ages of 10 and 15.

To clarify what I want:

Not wanted: ${e}^{i\pi}+1=0$ is cool, but (as far as I know) it can't be used for practical applications outside a classroom.

What I do want: e I think is used in calculations regarding compound interest. I'd like a simple explanation of how it is used (or links to simple explanations), and more examples like this.

How can e be used for practical mathematics? This is for a presentation on (among other numbers) e, aimed at people between the ages of 10 and 15.

To clarify what I want:

Not wanted: ${e}^{i\pi}+1=0$ is cool, but (as far as I know) it can't be used for practical applications outside a classroom.

What I do want: e I think is used in calculations regarding compound interest. I'd like a simple explanation of how it is used (or links to simple explanations), and more examples like this.

Algebra IAnswered question

beobachtereb 2022-09-18

A Bacteria Culture Contains 100 Cells and Grows at a Rate Proportional to its Size

A bacteria culture contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 420.

a) Find and expression for the number of bacteria after t hours

$P(t)=P(0){e}^{kt}=100$

b) Find the number of bacteria after 3 hours.

$P(t)=P(3){e}^{k3}=$ (confused as to how to set this up)

d) When will the population reach 10,000?

A bacteria culture contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 420.

a) Find and expression for the number of bacteria after t hours

$P(t)=P(0){e}^{kt}=100$

b) Find the number of bacteria after 3 hours.

$P(t)=P(3){e}^{k3}=$ (confused as to how to set this up)

d) When will the population reach 10,000?

Algebra IAnswered question

mksfmasterio 2022-09-17

Use the fact that the world population was 2560 million in 1950 and 3040 million in 1960

Use the fact that the world population was 2560 million in 1950 and 3040 million in 1960 to model the population of the world in the second half of the 20th century. (Assume that the growth rate is proportional to the population size.) What is the relative growth rate? Use the model to estimate the world population in 1993 and to predict the population in year 2020.

Solution We measure the time t in years and let t=0 in the year 1950. We measure the population P(t) in millions of people.

$P(0)=3040$

$P(t)=P(0){e}^{kt}=2560{e}^{kt}$

$P(10)=2560{e}^{kt}=3040$

This is where I have a problem.

I apply the natural logarithm to both sides of the equation.

$\mathrm{ln}2560{e}^{10k}=\mathrm{ln}3040$

I move the exponent up front. I am not sure if I am allowed to move the constant and variable.

$10k\text{}\mathrm{ln}2560e=\mathrm{ln}3040$

Use the fact that the world population was 2560 million in 1950 and 3040 million in 1960 to model the population of the world in the second half of the 20th century. (Assume that the growth rate is proportional to the population size.) What is the relative growth rate? Use the model to estimate the world population in 1993 and to predict the population in year 2020.

Solution We measure the time t in years and let t=0 in the year 1950. We measure the population P(t) in millions of people.

$P(0)=3040$

$P(t)=P(0){e}^{kt}=2560{e}^{kt}$

$P(10)=2560{e}^{kt}=3040$

This is where I have a problem.

I apply the natural logarithm to both sides of the equation.

$\mathrm{ln}2560{e}^{10k}=\mathrm{ln}3040$

I move the exponent up front. I am not sure if I am allowed to move the constant and variable.

$10k\text{}\mathrm{ln}2560e=\mathrm{ln}3040$

Algebra IAnswered question

mydaruma25 2022-09-17

An asymptotic behavior of ${\mathrm{Li}}_{-n}(a)$ for $n\to \mathrm{\infty}$

Suppose $a,b\in (0,1)$. I'm interested in comparison of an asymptotic behavior of ${\mathrm{Li}}_{-n}(a)$ and ${\mathrm{Li}}_{-n}(b)$ for $n\to \mathrm{\infty}$.

Such functions exhibit approximately factorial-like (faster than exponential) growth rate. The particular case ${\mathrm{Li}}_{-n}\phantom{\rule{negativethinmathspace}{0ex}}\left({\textstyle \frac{1}{2}}\right)$ for $n\ge 1$ gives (up to a coefficient) a combinatorial sequence called Fubini numbers or orderded Bell numbers[1][2][3] (number of outcomes of a horse race provided that ties are possible). This sequence is known to have the following asymptotic behavior:

$$\begin{array}{}\text{(1)}& {\mathrm{Li}}_{-n}\phantom{\rule{negativethinmathspace}{0ex}}\left({\textstyle \frac{1}{2}}\right)\sim \frac{n!}{{\mathrm{ln}}^{n+1}2}.\end{array}$$

After some numerical exprerimentation I conjectured the following behavior:

$$\begin{array}{}\text{(2)}& \mathrm{ln}\phantom{\rule{negativethinmathspace}{0ex}}\left(\frac{{\mathrm{Li}}_{-n}(a)}{{\mathrm{Li}}_{-n}(b)}\right)=(n+1)\cdot \mathrm{ln}\phantom{\rule{negativethinmathspace}{0ex}}\left(\frac{\mathrm{ln}b}{\mathrm{ln}a}\right)+o\phantom{\rule{negativethinmathspace}{0ex}}\left({n}^{-N}\right)\end{array}$$

for arbitrarily large N (so, the remainder term decays faster than any negative power of n). It looks like the remainder term is oscillating with exponentially decreasing amplitude, but I haven't yet found the exact exponent base or asymptotic oscillation frequency.

Could you suggest a proof of (2) or further refinements of this formula?

Suppose $a,b\in (0,1)$. I'm interested in comparison of an asymptotic behavior of ${\mathrm{Li}}_{-n}(a)$ and ${\mathrm{Li}}_{-n}(b)$ for $n\to \mathrm{\infty}$.

Such functions exhibit approximately factorial-like (faster than exponential) growth rate. The particular case ${\mathrm{Li}}_{-n}\phantom{\rule{negativethinmathspace}{0ex}}\left({\textstyle \frac{1}{2}}\right)$ for $n\ge 1$ gives (up to a coefficient) a combinatorial sequence called Fubini numbers or orderded Bell numbers[1][2][3] (number of outcomes of a horse race provided that ties are possible). This sequence is known to have the following asymptotic behavior:

$$\begin{array}{}\text{(1)}& {\mathrm{Li}}_{-n}\phantom{\rule{negativethinmathspace}{0ex}}\left({\textstyle \frac{1}{2}}\right)\sim \frac{n!}{{\mathrm{ln}}^{n+1}2}.\end{array}$$

After some numerical exprerimentation I conjectured the following behavior:

$$\begin{array}{}\text{(2)}& \mathrm{ln}\phantom{\rule{negativethinmathspace}{0ex}}\left(\frac{{\mathrm{Li}}_{-n}(a)}{{\mathrm{Li}}_{-n}(b)}\right)=(n+1)\cdot \mathrm{ln}\phantom{\rule{negativethinmathspace}{0ex}}\left(\frac{\mathrm{ln}b}{\mathrm{ln}a}\right)+o\phantom{\rule{negativethinmathspace}{0ex}}\left({n}^{-N}\right)\end{array}$$

for arbitrarily large N (so, the remainder term decays faster than any negative power of n). It looks like the remainder term is oscillating with exponentially decreasing amplitude, but I haven't yet found the exact exponent base or asymptotic oscillation frequency.

Could you suggest a proof of (2) or further refinements of this formula?

Algebra IAnswered question

Keenan Conway 2022-09-17

Solution for differential equation describing linear input and exponential decay

I try to model the concentration over time when there's an linear input and an exponential output (i.e. exponential decay) with known half-life.

Input(t) = a+b*t

where t is the time (in years), a the input at year 0 and b the yearly growth to that input.

Output(t) = lambda*N(t)

where t is the time (in years), lambda the exponential decay constant and N(t) the concentration at year t.

lambda is defined as

lambda = ln(2)/th

where th is the half-life in years.

The differential equation is therefore:

dN/dt = Input - Output = a+b*t-lambda*N(t)

Using wolframAlpha I get this solution:

N(t) = C*exp(-lambda*t) + (b/lambda)*t + (a/lambda) - (b/lambda^2)

where C is the initial concentration at t=0.

For small values of th this makes sense, but if th assumes higher values, the concentration is negative. How can this be?

As example:

a = 300

b = 20

th = 2

C = 300

For t = 10 I get a concentration of 1285.561.

But if I set th to 200, I get -1520536 at t = 10

I try to model the concentration over time when there's an linear input and an exponential output (i.e. exponential decay) with known half-life.

Input(t) = a+b*t

where t is the time (in years), a the input at year 0 and b the yearly growth to that input.

Output(t) = lambda*N(t)

where t is the time (in years), lambda the exponential decay constant and N(t) the concentration at year t.

lambda is defined as

lambda = ln(2)/th

where th is the half-life in years.

The differential equation is therefore:

dN/dt = Input - Output = a+b*t-lambda*N(t)

Using wolframAlpha I get this solution:

N(t) = C*exp(-lambda*t) + (b/lambda)*t + (a/lambda) - (b/lambda^2)

where C is the initial concentration at t=0.

For small values of th this makes sense, but if th assumes higher values, the concentration is negative. How can this be?

As example:

a = 300

b = 20

th = 2

C = 300

For t = 10 I get a concentration of 1285.561.

But if I set th to 200, I get -1520536 at t = 10

Algebra IAnswered question

Hrefnui9 2022-09-14

"When zombies finally take over, the population of the Earth will decrease exponentially. Every HOUR that goes by the human population will decrease by 5%. The population today is 6,000,000,000, find a function P that gives the population of the earth d DAYS after the beginning of the zombie takeover."

This without a doubt puts me in between a rock and a difficult place. As you may consider I researched this query and numerous methods of exponential boom everywhere on line, in my textbook, and in my class notes, but I nonetheless come complete circle and locate myself lower back to square one, and not using a clue what so ever on how this works. i might really admire it if someone should at least factor me inside the right direction like the way to version a decay with exponential variables regarding HOURS and DAYS because the query says. I already recognize my preliminary method being 6,000,000,000(0.85) I simply need to know the exponents used! Is it 6,000,000,000(0.eighty five)^d/60, 6,000,000,000(zero.eighty five)^d/1? Or some thing like that? Please assist I actually need to understand this idea!

This without a doubt puts me in between a rock and a difficult place. As you may consider I researched this query and numerous methods of exponential boom everywhere on line, in my textbook, and in my class notes, but I nonetheless come complete circle and locate myself lower back to square one, and not using a clue what so ever on how this works. i might really admire it if someone should at least factor me inside the right direction like the way to version a decay with exponential variables regarding HOURS and DAYS because the query says. I already recognize my preliminary method being 6,000,000,000(0.85) I simply need to know the exponents used! Is it 6,000,000,000(0.eighty five)^d/60, 6,000,000,000(zero.eighty five)^d/1? Or some thing like that? Please assist I actually need to understand this idea!

Algebra IAnswered question

frobirrimupyx 2022-09-14

Why is e so special?

The number e (and the exponentiation function ${e}^{x}$) appears in so many places in mathematics and engineering. There seem to be a multitude of applications of it. I want to know why.

The number e (and the exponentiation function ${e}^{x}$) appears in so many places in mathematics and engineering. There seem to be a multitude of applications of it. I want to know why.

Algebra IAnswered question

Staffangz 2022-09-14

$y=a(1+r{)}^{t}$

I know this is a really basic question for this website, but I can't find it anywhere else.

This is the question: "If you deposit $3,750 in an account that pays 6% annual interest compounded monthly, what is the balance of the account after 11 years?"

The formula I'm using is $y=a(1+r{)}^{t}$, with a being the initial amount, r being the rate in decimal form, and t is time relative to the rate, which makes $y=3,750(1+.06{)}^{132}$

How do I solve for the ending amount (y)?

I know this is a really basic question for this website, but I can't find it anywhere else.

This is the question: "If you deposit $3,750 in an account that pays 6% annual interest compounded monthly, what is the balance of the account after 11 years?"

The formula I'm using is $y=a(1+r{)}^{t}$, with a being the initial amount, r being the rate in decimal form, and t is time relative to the rate, which makes $y=3,750(1+.06{)}^{132}$

How do I solve for the ending amount (y)?

Algebra IAnswered question

blogswput 2022-09-14

How do Mathematicians determine when to use the constant e ?

I'm only an undergrad so sorry if this is a dumb question, but I was studying Poisson distribution and it struck me that so many models involve "e". So it got me wondering; how/when/where/why do they decide to use it? I'm assuming they don't build a model and include "e" just because, and there must be some sort of fundamental intuition behind when its use is appropriate.

I'm only an undergrad so sorry if this is a dumb question, but I was studying Poisson distribution and it struck me that so many models involve "e". So it got me wondering; how/when/where/why do they decide to use it? I'm assuming they don't build a model and include "e" just because, and there must be some sort of fundamental intuition behind when its use is appropriate.

Algebra IAnswered question

blogswput 2022-09-12

A Bacteria Culture Grows with Constant Relative Growth Rate.

A micro organism culture Grows with consistent Relative boom rate. The micro organism count number become 400 after 2 hours and 25,six hundred after 6 hours.

a) What is the relative growth rate? Express your answer as a percentage.

Using this formula $P(t)={P}_{0}{e}^{kt}$

$t=time$

$k=growth\text{}rate$

${p}_{0}=initial\text{}amount$

How do I calculate a formula with the information I have?

b) What was the initial size of the culture?

c) Find an expression for the number of bacteria after t hours

d) Find the number of cells after 4.5 hours

e) Find the rate of growth after 4.5 hours

f) When the population reach 50,000?

that is a self look at query. i've Calculus II subsequent semester and sincerely would like to have a clear method as to a way to solve a hassle inclusive of this.

A micro organism culture Grows with consistent Relative boom rate. The micro organism count number become 400 after 2 hours and 25,six hundred after 6 hours.

a) What is the relative growth rate? Express your answer as a percentage.

Using this formula $P(t)={P}_{0}{e}^{kt}$

$t=time$

$k=growth\text{}rate$

${p}_{0}=initial\text{}amount$

How do I calculate a formula with the information I have?

b) What was the initial size of the culture?

c) Find an expression for the number of bacteria after t hours

d) Find the number of cells after 4.5 hours

e) Find the rate of growth after 4.5 hours

f) When the population reach 50,000?

that is a self look at query. i've Calculus II subsequent semester and sincerely would like to have a clear method as to a way to solve a hassle inclusive of this.

Algebra IAnswered question

ubwicanyil5 2022-09-12

A town doubles its populace in 25 years. If it's far growing exponentially, while will it triple its population?

The above is a query in my maths textbook within the subject matter Exponential increase & Decay.

i am a piece confused as to how I have to technique this query.

We have been taught to use the formula:

$$Q=A{e}^{kt}$$

Where Q is the quantity, A is the initial quantity, k is the growth/decay constant and t is the time.

In reference to the question, I don't think I need A so here is the equation I ended up with:

$$2Q={e}^{25k}$$

Edit:

I found out that

$$k=\frac{\mathrm{ln}2}{25}$$

I then let Q=3A and the following is my working:

$$3A=A{e}^{25\frac{\mathrm{ln}2}{25}t}$$

$$3A=A{e}^{\mathrm{ln}2t}$$

$$3={e}^{\mathrm{ln}2t}$$

$$3={2}^{t}$$

$$\mathrm{ln}3=t\mathrm{ln}2$$

$$t=\frac{\mathrm{ln}3}{\mathrm{ln}2}$$

$$t=1.6$$

I can't figure out what is wrong in my working out.

The provided answer is: 39.6 years

The above is a query in my maths textbook within the subject matter Exponential increase & Decay.

i am a piece confused as to how I have to technique this query.

We have been taught to use the formula:

$$Q=A{e}^{kt}$$

Where Q is the quantity, A is the initial quantity, k is the growth/decay constant and t is the time.

In reference to the question, I don't think I need A so here is the equation I ended up with:

$$2Q={e}^{25k}$$

Edit:

I found out that

$$k=\frac{\mathrm{ln}2}{25}$$

I then let Q=3A and the following is my working:

$$3A=A{e}^{25\frac{\mathrm{ln}2}{25}t}$$

$$3A=A{e}^{\mathrm{ln}2t}$$

$$3={e}^{\mathrm{ln}2t}$$

$$3={2}^{t}$$

$$\mathrm{ln}3=t\mathrm{ln}2$$

$$t=\frac{\mathrm{ln}3}{\mathrm{ln}2}$$

$$t=1.6$$

I can't figure out what is wrong in my working out.

The provided answer is: 39.6 years

Algebra IAnswered question

Slovenujozk 2022-09-12

How does exp(x) keep showing up in mathematics

When I first learnt about e, I just treated it as another number (as far as I know, it doesn't even have a natural definition), but how is it that exp(x) is so important and keeps showing up at various places in mathematics?

When I first learnt about e, I just treated it as another number (as far as I know, it doesn't even have a natural definition), but how is it that exp(x) is so important and keeps showing up at various places in mathematics?

Algebra IAnswered question

tuzkutimonq4 2022-09-11

Consider the single species population model defined by

$$\frac{dR}{dt}=\frac{gR}{R+{R}_{m}}-dR,$$

for t>0, where $g,{R}_{m}$, and d are all positive parameters and $R(0)={R}_{0}$

(a) Describe the biological meaning of each term in the equation.

(b) Determine the steady-states of the system and discuss any constraints on the model parameters for the model to admit biologically meaningful solutions.

(c) Determine the steady-state stability and discuss any variation in this with respect to the model parameter values.

a) gR represents the exponential growth of population, dR represents the exponential decay of population, g is the growth rate, and d is the decay rate. What do R and ${R}_{m}$ represent? How can I define the term

$$\frac{gR}{R+{R}_{m}}?$$

What is $R+{R}_{m}$? Does it affect the gR for the grow?

b) In single steady-state system,

$$\frac{dR}{dt}=0.$$

$$\frac{gR}{R+{R}_{m}}-dR=0$$

$$gR-dR(R+{R}_{m})=0$$

$$R(g-d(R+{R}_{m}))=0$$

either R=0 or $g-d(R+{R}_{m})=0$

$$g-d{R}_{m}=0\text{}\text{}\text{}(R=0)$$

$${R}_{m}=\frac{g}{d}$$

$${R}^{\ast}=\frac{g}{d}\text{}\text{}\text{}({R}_{m}={R}^{\ast})$$

So, we have ${R}_{1}^{ast}=0$,${R}_{2}^{\ast}=\frac{g}{d}$ Are these correct? I am not sure constraint on the model parameters to admit biologically meaning solutions?

c) to determine steady-state stability let

$$f(R)=\frac{gR}{R+{R}_{m}}-dR$$

$$\frac{df}{dR}=gln(R+{R}_{m})-d.$$

My differentiation may be wrong and don't know the term ${R}_{m}$ while differentiating with respect to R. I really don't know after that. and I know my answer is still incomplete.

$$\frac{dR}{dt}=\frac{gR}{R+{R}_{m}}-dR,$$

for t>0, where $g,{R}_{m}$, and d are all positive parameters and $R(0)={R}_{0}$

(a) Describe the biological meaning of each term in the equation.

(b) Determine the steady-states of the system and discuss any constraints on the model parameters for the model to admit biologically meaningful solutions.

(c) Determine the steady-state stability and discuss any variation in this with respect to the model parameter values.

a) gR represents the exponential growth of population, dR represents the exponential decay of population, g is the growth rate, and d is the decay rate. What do R and ${R}_{m}$ represent? How can I define the term

$$\frac{gR}{R+{R}_{m}}?$$

What is $R+{R}_{m}$? Does it affect the gR for the grow?

b) In single steady-state system,

$$\frac{dR}{dt}=0.$$

$$\frac{gR}{R+{R}_{m}}-dR=0$$

$$gR-dR(R+{R}_{m})=0$$

$$R(g-d(R+{R}_{m}))=0$$

either R=0 or $g-d(R+{R}_{m})=0$

$$g-d{R}_{m}=0\text{}\text{}\text{}(R=0)$$

$${R}_{m}=\frac{g}{d}$$

$${R}^{\ast}=\frac{g}{d}\text{}\text{}\text{}({R}_{m}={R}^{\ast})$$

So, we have ${R}_{1}^{ast}=0$,${R}_{2}^{\ast}=\frac{g}{d}$ Are these correct? I am not sure constraint on the model parameters to admit biologically meaning solutions?

c) to determine steady-state stability let

$$f(R)=\frac{gR}{R+{R}_{m}}-dR$$

$$\frac{df}{dR}=gln(R+{R}_{m})-d.$$

My differentiation may be wrong and don't know the term ${R}_{m}$ while differentiating with respect to R. I really don't know after that. and I know my answer is still incomplete.

Exponential growth and decay subject related to one of the more complex aspects of Algebra, which makes it relatively difficult for students to cope with it as it requires analysis and knowledge of the basics. Take your time to explore various exponential growth and decay practice answers below to refresh your memory and see some helpful examples.

The answers that you can see below must be linked to the questions to see the reasons why certain solutions have been provided. Remember that analysis will be helpful to see the correct approach!