Expert Answers to Algebra 2 Problems

Solve
$8(y-<!--\; -\; -->4)=5(y+4)$

Am I allowed to apply L'Hospital's Rule inside of the natural logarithm function?
I have the following limit:
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\ln <!--\; \; -->\left(\frac{2x^2+1}{x^2+1}\right)$
If I was finding the limit of only the terms inside the natural log function, I would have the indeterminate form:
$\frac{\mathrm{\infty <!--\; \infty \; -->}}{\mathrm{\infty <!--\; \infty \; -->}}$
I want to know if I am allowed to apply L'Hospital's Rule to only the inside terms while ignoring the natural logarithm function, giving me the answer:
$\ln <!--\; \; -->\left(\frac{4x}{2x}\right)=\ln <!--\; \; -->(2)$

$\underset{x=a}{\overset{b-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}\frac{1}{x}$ and $\underset{x=a+1}{\overset{b}{\sum <!--\; \sum \; -->}}\frac{1}{x}$
I have to prove the following relations:
$\underset{x=a}{\overset{b-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}\frac{1}{x}\ge <!--\; \ge \; -->\log <!--\; \; -->b-<!--\; -\; -->\log <!--\; \; -->a$
$\underset{x=a+1}{\overset{b}{\sum <!--\; \sum \; -->}}\frac{1}{x}\le <!--\; \le \; -->\log <!--\; \; -->b-<!--\; -\; -->\log <!--\; \; -->a$
I tried to use the relation that ${\int <!--\; \int \; -->}_a^b\frac{1}{x}dx=\log <!--\; \; -->b-<!--\; -\; -->\log <!--\; \; -->a$ and for the second relation I obtained that ${\int <!--\; \int \; -->}_{a+1}^b\frac{1}{x}dx=\log <!--\; \; -->b-<!--\; -\; -->\log <!--\; \; -->(a+1)<\log <!--\; \; -->b-<!--\; -\; -->\log <!--\; \; -->a$ and I think that could be sufficient to prove the relation. Instead, for the first relation i have some problems, because using the same method I don't obtain a good result, indeed: ${\int <!--\; \int \; -->}_a^{b-<!--\; -\; -->1}\frac{1}{x}dx=\log <!--\; \; -->(b-<!--\; -\; -->1)-<!--\; -\; -->\log <!--\; \; -->a$ that is not greater than $\log <!--\; \; -->b-<!--\; -\; -->\log <!--\; \; -->a$. What do you think? How can I resolve it?

Why does this equation work?
let $P(x):=\underset{p\le <!--\; \le \; -->x}{\sum <!--\; \sum \; -->}Log[p]$, then we have
$P(2^{k+1})=\underset{i=0}{\overset{k}{\sum <!--\; \sum \; -->}}(P(2^{i+1})-<!--\; -\; -->P(2^i))<2\cdot <!--\; \cdot \; -->Log[2]\cdot <!--\; \cdot \; -->(1+2+4+...+2^k)\le <!--\; \le \; -->4\cdot <!--\; \cdot \; -->Log[2]\cdot <!--\; \cdot \; -->2^k$
Why does this equation work?
The first step is clear for me, since we have a telescope sum. However, how can I conclude to the first $<$ and the $\le <!--\; \le \; -->$ ?

What is the derivation for the derivative of $a^t$
Been driving me nuts. Can someone prove to me that
$\frac{d(a^t)}{dt}=a^t\ln <!--\; \; -->(a)$
Thank you!

How can $\frac{1}{2}\log <!--\; \; -->(x)=\log <!--\; \; -->(\sqrt{x})$?
How can $\frac{\log <!--\; \; -->(x)}{2}=\log <!--\; \; -->\left(\sqrt{x}\right)$?
How would I come to this conclusion?

How to simplify ${\ln }^2<!--\; \; -->\left(x\right)+2\ln <!--\; \; -->x-<!--\; -\; -->3$
I dont know how to simplify ${\ln }^2<!--\; \; -->\left(x\right)+2\ln <!--\; \; -->x-<!--\; -\; -->3$
I dont know how to get $(\ln <!--\; \; -->(x)+1)(\ln <!--\; \; -->(x)+3)$
But I am stuck and don't really know how to do that.
I tried something like this: $2\ln <!--\; \; -->x+2\ln <!--\; \; -->x-<!--\; -\; -->3,2\ln <!--\; \; -->\left(x(x-<!--\; -\; -->3)\right),2\ln <!--\; \; -->(x^2-<!--\; -\; -->3x)$

Tighter logarithmic inequality
There is a well-known lower bound for
$x\log <!--\; \; -->\frac{1+x}{x}\ge <!--\; \ge \; -->\frac{x}{1+x}$
for $x\ge <!--\; \ge \; -->0$. I know a tighter lower bound on the same domain
$x\log <!--\; \; -->\frac{1+x}{x}\ge <!--\; \ge \; -->\frac{2x}{1+2x}\ge <!--\; \ge \; -->\frac{x}{1+x}.$
It can be proved by Jensen's inequality. Do you know something tighter?

If 1 kg = 2.2 lbs
How many pounds equal 120 kilograms?
lbs

A logarithm integral
Calculate the integral
$\begin{array}{r}{\int <!--\; \int \; -->}_0^1\frac{\ln <!--\; \; -->(\sqrt{x}-<!--\; -\; -->\sqrt{1-<!--\; -\; -->x})}{\sqrt{x}}dx\end{array}$
and show the value is negative.

A ball is kicked with an initial velocity of 19.0 m/s in the horizontal direction and 17.0 m/s in the vertical direction. (Assume the ball is kicked from the ground.)
(a) At what speed in m/s) does the ball hit the ground? m/s
(b) For how long in s) does the ball remain in the air?
(c) What maximum height (in m) is attained by the ball?

Find a polynomial f(x) of degree 5 that has the following zeros.
-9, 5, -7, 4, 8
f(x)=

When x = a is a zero of a polynomial function f, the three statements below are true.
(a) x = a is a ________ of the polynomial equation f(x) = 0.
(b) ________ is a factor of the polynomial f(x).
(c) (a, 0) is an ________ of the graph of f.

Simplifying/solving a logarithm ${\log }_2<!--\; \; -->4^{2n}$
Need help with simplifying this logarithm.
${\log }_2<!--\; \; -->4^{2n}$
Would I just pull the 2n to the front:
$2n\ast <!--\; \ast \; -->{\log }_2<!--\; \; -->4$
So it would simplify to
$4n$
Is this correct or am I completely wrong?

Derivative of $f(x{)}^{g(x)}$ at points when $f(x)=0$
I am interested in understanding the general behavior of the derivative for
$f(x{)}^{g(x)}$
at points where $f(x)=0$
For example, if $f^g=x^n$ we have
$\frac{d}{dx}{f}^g(0)=\{\begin{array}{ll}0& n\ge <!--\; \ge \; -->1\\ \pm <!--\; \pm \; -->\mathrm{\infty <!--\; \infty \; -->}& n<1\end{array}$
The general formula
$(f^g{)}^{\prime }=f^g(g^{\prime }\ln <!--\; \; -->|f|+g\frac{f^{\prime }}{f})$
breaks down when $f=0$, though as the example above shows the derivative may still exist there.
I am not sure what the proper assumptions should be. Tentatively, take $f(x{)}^{g(x)}\ge <!--\; \ge \; -->0$ for all $x$ in the relevant domain, so that (I believe) we have
$\ln <!--\; \; -->(f(x{)}^{g(x)})=g(x)\ln <!--\; \; -->|f(x)|$
when $f$ is strictly positive and undefined otherwise.
Also, is there any non-trivial example of a well-defined function (meaning "nice", as in differentiable almost everywhere) $f^g$ where $f$ takes on negative values and where $g$ is not constant? In other words, for cases of $f<0$ are the only functions worth considering of the form $f^c$ for constants $c$?
EDIT: I want the function(s) to be real, although arguments using complex numbers are of course permissible.

solve for $x$ without using softwares ${\log }_{\sqrt{x}}<!--\; \; -->2+{\log }_6<!--\; \; -->x^x=4$
Is there any nice way to solve this equation without wolfram?
${\log }_{\sqrt{x}}<!--\; \; -->2+{\log }_6<!--\; \; -->x^x=4$
Thanks.

Solving the system with logarithms
I tried solving the system $\{\begin{array}{l}(4x{)}^{{\log }_2<!--\; \; -->(2y)}=64\\ (8y{)}^{{\log }_2<!--\; \; -->(2y)}=256\end{array}$ several times but still keep getting wrong solutions.

Prove equality $a^{{\log }_b<!--\; \; -->c}=c^{{\log }_b<!--\; \; -->a}$
I'm try to prove the equality:
$a^{{\log }_b<!--\; \; -->c}=c^{{\log }_b<!--\; \; -->a}$
I'm having trouble finding information regarding this, also I need to figure out why $n^{{\log }_2<!--\; \; -->3}$ is better than $3^{{\log }_2<!--\; \; -->n}$ as a closed form formula for $M(n)$? Thank you!

Limit of a Logarithm with Different Bases
We are to compute
$\underset{n-<!--\; -\; -->>\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{2^{{\log }_3<!--\; \; -->n}}{3^{{\log }_2<!--\; \; -->n}}$
Clearly the bases are reversed between the logarithm and exponents, so I can't seem to find any logarithm or exponential properties that can help simplify this problem.

The data sources and methods for estimating the costs, benefits and disbenefits for the nuclear power plant.