# Exponential Growth & Decay Equations & Examples

Recent questions in Exponential growth and decay
Ronin Daniels 2023-03-29

## In how many different orders can five runners finish a race if no ties are allowed???

Algebra IOpen question
deal4markaz 2022-12-06

## Q:   Find the product with the exponent in simplest form. Then, identify the values of x and y. 6^1/3x6^1/4=6^x/y X= ,y=

Algebra IOpen question
deal4markaz 2022-12-06

## Q:   Find the product with the exponent in simplest form. Then, identify the values of x and y. 6^1/3x6^1/4=6^x/y X= ,y=

arponatsdWT 2022-11-30

## Natalie's Nail Salon has 2 locations: Natalie's Downtown, and Natalie's By-the-Beach. The two salons combined performed 320 pedicures in the first quarter of 2011, and 400 pedicures in the second quarter. Pedicures cost \$8.50 at both locations in Q1 and Q2. Natalie notices that in Q1, 30% of pedicures are performed at By-the-Beach but in Q2, 55% of the total were performed there. Natalie's sales goals for Quarter 3 are 6% growth in each location compared to Quarter 2. Natalie is considering raising pedicures prices to \$11 at both locations in Q3 to increase revenue.What was the overall pedicure growth rate from Q1 to Q2?

Ayanna Goodman 2022-11-20

## Why is e used in exponential growth functions?Now I'm aware that ${e}^{x}$ is its own derivative, which makes it convenient to use in calculus. However, I have a question about this function:Intuitively, an exponential growth function could be written as $a\ast \left(1+k{\right)}^{t}$, where a is the initial amount, k is the growth rate and t is the time.However, it can also be rewritten as $a\ast {e}^{kt}$, same thing, but with Euler's number added to it.In what way are these both the same? They yield the same answer. Why not 2, or 4, or 10? I know that one would need to change the growth constant if using one of those bases, the one I called k, but how come one doesn't need to change it when it's in base e?

paratusojitos0yx 2022-11-17

## Doing some computations and plottings I've found out that the function${}_{2}{F}_{1}\left(2s-1,s-\frac{1}{2};s;-1\right)$behaves for large real s like ${4}^{-s}$. More precisely: It seems that${}_{2}{F}_{1}\left(2s-1,s-\frac{1}{2};s;-1\right){4}^{s}$grows, but very very slowly. No exponential growth or decay at all. If you modify the 4 only slightly this of course changes and you get exponential groth or decay. So my question is to explain this phenomenon. IsIs there an easy to detemine for every x>0 the respective b(x)>0 such that you have${}_{2}{F}_{1}\left(2s-1,s-\frac{1}{2};s;-x\right)\sim {b}^{-s}?$What isthen?

Arendrogfkl 2022-11-16

## I'm interested in solving a heat equation with a discontinuous source term in one dimension on the real line:$\frac{\mathrm{\partial }u}{\mathrm{\partial }t}\left(x,t\right)=-x\theta \left(x\right)u\left(x,t\right)-x\theta \left(-x\right)u\left(-x,t\right)+D\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{x}^{2}}\left(x,t\right)$where $\theta \left(\cdot \right)$ is the Heaviside function. The boundary conditions are that $\underset{|x|\to \mathrm{\infty }}{lim}u\left(x,t\right)=0$ The initial condition is arbitrary, but even for a symmetric Gaussian or a symmetric exponential an analytical solution would be interesting.Because of the discontinuity in the right-hand side induced by $\theta \left(\cdot \right)$, my intuition would be to solve in the two domains x>0 and x<0 and patch them together at x=0. However, the source terms have the effect of causing exponential decay with rate x at the point x for x>0 and corresponding growth at the point −x. This introduces an asymmetry, which should cause a diffusive flux back to the side x>0. Solving for the solution in the two domains does not seem to capture this feedback, and I'm not sure how to solve globally. What are some avenues to proceed?

Amy Bright 2022-11-10

## If ${\int }_{|x|\le r}|f\left(x\right)|dx\le \left(r+1{\right)}^{a}$, then ${\int }_{\mathbb{R}}|f\left(x\right)|{e}^{-|tx|}dx<\mathrm{\infty }$ for $t\ne 0$Let f be some function in ${L}_{loc}^{1}\left(\mathbb{R}\right)$ such that, for some $a\in \mathbb{R}$${\int }_{|x|\le r}|f\left(x\right)|dx\le \left(r+1{\right)}^{a}$for all $r\ge 0$. Show that $f\left(x\right){e}^{-|tx|}\in {L}^{1}\left(\mathbb{R}\right)$ for all $t\in \mathbb{R}\setminus \left\{0\right\}$I'm having a hard time finding use of the bound described above. Any help would be appreciated.

Abdiel Mays 2022-11-08

## I want to know why the equation ${y}^{2}=1-\frac{4{x}^{{10}^{12}}}{{\pi }^{2}}$ gives an approximate square.BackgroundI was just playing around with functions and I wanted to see if $y=|\mathrm{sin}\left(\frac{\pi x}{2}\right)|$ (radians) would give a semicircle for the interval [0,2] as the distance of (1,0) is the same from (0,0), (2,0) and (1,1), all of which will lie on the curve. The equation of a unit semicircle with its centre at (1,0) is $y=\sqrt{2x-{x}^{2}}$I know that the curves of both the equations don't resemble each other much but I still thought of approximating the sine function using this because I thought that it could still be combined with another approximation to make a better approximation. Anyway, I did it and for , the value of sinϕ can to be approximately $\frac{2}{\pi }\sqrt{\pi x-{x}^{2}}$. It looked like a semi-ellipse and so I verified it to find that it was a semi-ellipse. I thought of using this to derive the equation for an ellipse with it's centre at the origin and the value of $a$ and b being $\frac{\pi }{2}$ and 1 respectively.The equation came out to be : ${y}^{2}=1-\frac{4{x}^{{10}^{12}}}{{\pi }^{2}}$Finally, I thought of playing with this equation and changed the exponent of x. I observed that as I increased the power, keeping it even, the figure got closer and closer to a square.${y}^{2}=1-\frac{4{x}^{{10}^{12}}}{{\pi }^{2}}$ gave a good approximation of a square. For the exponent of x being some power of 10 greater than 1012, a part of the curve began to disappear.

evitagimm9h 2022-11-06

## ${e}^{pt}$or $\left(1+p{\right)}^{t}$ What is the difference in modeling exponential growth and decay?I would really like to better recognize, whilst the feature ${e}^{pt}$ is the "higher" desire and while (if at all) $\left(1+p{\right)}^{t}$ have to be used.to provide a conventional example: Say we need to version radioactive decay of some detail A. allow t be in units of one half-life of A. Then $p=-\frac{1}{2}$Now I'd say the standard approach to modeling this is via the function${A}_{1}\left(t\right)={A}_{0}\cdot {\left(1-\frac{1}{2}\right)}^{t}.$On the other hand, if we approach this problem as an ODE, we can say that at any point t we want A(t) to decrease at a rate of half of its momentary amount:which leads to the function${A}_{2}\left(t\right)={A}_{0}\cdot {e}^{-\frac{1}{2}t}.$But which approach would be "better" here? I think ${A}_{1}$ is much more commonly (if not exceptionally) used when it comes to modeling atomic decay. On the other hand, I know that$e=\underset{n\to \mathrm{\infty }}{lim}{\left(1+\frac{1}{n}\right)}^{n}$which essentially means that the rate of change of ${A}_{2}$ is continously updated, while ${A}_{1}$ is updated discretely, right? That's the best way I can phrase it at the moment.So to conclude: Does this mean that ${A}_{2}$ is always the "better", more accurate choice or are there situations where ${A}_{1}$ is actually "correct"?

jorgejasso85xvx 2022-11-05

## Would $-{3}^{-x}$ be an exponential decay of growth? Any and all help appreciated.

Aliyah Thompson 2022-11-04

## Stability of Finite Difference schemes.I was reading a book ''Numerical partial differential equation: Finite difference methods'' by J. W. Thomas. Here at page no. 74, definition of stability is given as$‖{u}^{n+1}‖\le K{e}^{\beta t}‖{u}^{0}‖$ for $0\le \mathrm{\Delta }x\le \mathrm{\Delta }{x}_{0}$ and $0\le \mathrm{\Delta }t\le \mathrm{\Delta }{t}_{0}$. i.e. this definition allows for exponential growth. however in a few literature i found the definition that ''A finite distinction approximation is solid if the errors (truncation, spherical-off and so forth) decay as the computation proceeds from one marching step to the next .'' The ebook which i've noted, at the equal page in remark 3, creator has written that the above definition with exponential increase is trendy definition and later one may be derived from this.but i am not getting the way to locate this definition. So my question is which definition is true ? And if different one is not proper, then what is motive at the back of ? and the way can we examine those two definition?

Kenna Stanton 2022-11-02

## How would you best describe the rate of growth of the function $f\left(x\right)=cx{r}^{x}$ ?Consider the function $f\left(x\right)=cx{r}^{x}$ , where both r and c are constants and we have cases: (a) r<1, (b) r>1. Regarding terminology, how would you best describe the asymptotic growth of f(x) in cases (a) and (b)? Though you could state that exponential growth and decay will dominate the asymptotics of the function, strictly speaking it would be incorrect to call (a) exponential growth and (b) exponential decay, right?

Marilyn Cameron 2022-10-26

## Consider the modification to the Malthusian equation$\frac{dN}{dt}=rS\left(N\right)N,$where r>0 is the per capita growth rate, and S(N) is a survival fraction. For some organisms, finding a mate at low population densities may be difficult. In such cases, the survival fraction can take the form $S\left(N\right)=\frac{N}{A+N}$, where A>0 is a constant.(a)i. By examining what happens to S(N) for N≫A and N≪A explain why this fraction models the situation outlined above.Attempt:When N≫A (my assumption is that the sign '≫'means significantly greater than ) so A is much smaller than N then the constant A.This would effect the survival fraction in a way which it becomes roughly equal to 1 since A is really small.When N≪A (my assumption is that the sign '≪' means significantly smaller than) so N is much smaller than A will impact the survival fraction in a way which will make it become small.The survival fraction models matches with the situation above since:For a low population finding a mate is difficult (a.k.a N≪A )For a large population finding a mate is more likely (a.k.a N≫A )ii. By examining the form of the equation, determine the long-term behaviour of a population for an initial condition N(0)>0.Attempt: For the initial conditions stated above the Malthusian equation has the form:$\frac{dN}{dt}=r\phantom{\rule{thinmathspace}{0ex}}\frac{N}{A+N}\phantom{\rule{thinmathspace}{0ex}}N,$which has the solution.$N\left(t\right)={N}_{0}{e}^{\lambda t}$If $\lambda >0$ then exponential growthIf $\lambda <0$ then exponential decayPlease could you check if this is correct. If it is please can you suggest any improvement I could add to my attempts.

Francis Oliver 2022-10-22

## Why do we use the form $f\left(t\right)=a{e}^{kt}$ for exponential growth and decay?Why will we include the ${e}^{k}$?would not or not it's simpler to definitely use $f\left(t\right)=a{p}^{t}$ wherein p is the proportion growth according to time.Is there a reason why the convention is to apply $f\left(t\right)=a{e}^{kt}$?

Gardiolo0j 2022-10-01

## Experiments show that if the chemical reaction${\mathrm{N}}_{2}{\mathrm{O}}_{5}\to 2\phantom{\rule{thinmathspace}{0ex}}\mathrm{N}{\mathrm{O}}_{2}+\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}{\mathrm{O}}_{2}$takes place at $45{\phantom{\rule{thinmathspace}{0ex}}}^{\circ }\mathrm{C}$, the rate of the reaction of dinitrogen pentoxide is proportional to its concentration as follows:$-\frac{d\left[{\mathrm{N}}_{2}{\mathrm{O}}_{5}\right]}{dt}=0.0005\left[{\mathrm{N}}_{2}{\mathrm{O}}_{5}\right]$a) Find an expression for the concentration [${\mathrm{N}}_{2}{\mathrm{O}}_{5}$] after t seconds if the initial concentration is C.b) How long will the reaction take to reduce the concentration of ${\mathrm{N}}_{2}{\mathrm{O}}_{5}$ to 90% of its original value?Part aI simply used the instantaneous rate of 0.0005 as the constant k (referring to the rate of reaction) and, since C is a constant, I utilized it as the initial value in the formula:$y\left(t\right)=y\left(o\right){e}^{kt}=C{e}^{kt}$I replaced k for 0.0005: the book has a negative sign in front of this value, I am curious as to why since the equal sign would indicate otherwise, no? Nonetheless, the equation:$C{e}^{-0.0005\left(t\right)}$Please comment on my reasoning.Part bI simply utilized the above equation and set it equal to the decimal form of 90% with a negative sign since the question said reduced$C{e}^{-0.0005\left(t\right)}=-0.9$However, I really don't understand conceptually why 90% can be considered 0.9 of the concentration. I just need to clear this up in my mind.From that point I am pretty lost why the C appears on both sides of the equation and why it disappears.Answer from the book:$y\left(t\right)=C{e}^{-0.0005}=0.9C\to C{e}^{-0.0005}=0.9C\to {e}^{-0.0005}=0.9$What happened to C?

minuziavj 2022-09-30

## Units for rate of change, instantaneous or otherwiseIf we multiply a principal amount A0 in dollars (\$) by an interest rate r in percentage (%) once, let's say for a month (m), then we have an amount A in dollars per month (\$/m)$A\left(\text{/m}\right)={A}_{0}\left(\text{}\right)\cdot r\left(\text{%}\right)$The problem is that for the devices to agree arithmetically, the hobby rate might ought to be in units of in keeping with month (/m). that is additionally the case if we multiply the hobby rate by using a time frame t in months per year (m/y), then we've an amount in dollars consistent with yr(\$/y)$A\left(\text{/y}\right)={A}_{0}\left(\text{}\right)\cdot r\left(\text{%}\right)\cdot t\left(\text{m/y}\right)$Once again, the interest rate would have to be in units of per month (/m) which looks strange. This is still the case for the instantaneous rate of change of the amount with respect to time $\frac{dA}{dt}$ in dollars per period of time (\$/m), which looks a lot like the formula for exponential growth and decay$\frac{dA}{dt}\left(\text{/m}\right)={A}_{0}\left(\text{}\right)\cdot r\left(\text{%}\right)$Another time, the hobby fee could need to be in units of per month(/m). i have continually considered a charge like a scaling element with out wondering plenty of the units, so how am i able to make feel of the gadgets for the above interest charge?

tonan6e 2022-09-29