Expert Answers to Algebra 2 Problems

Integrating $\ln <!--\; \; -->(x)\times <!--\; \times \; -->\ln <!--\; \; -->(1-<!--\; -\; -->x)$
Is there a way I can derive the value of the integral
${\int <!--\; \int \; -->}_0^1\ln <!--\; \; -->(x)\ln <!--\; \; -->(1-<!--\; -\; -->x)dx$
using the fact that
${\displaystyle \underset{i=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{1}{n^2}=\frac{{\mathrm{\pi <!--\; \pi \; -->}}^2}{6}}$
? (the actual value of the integral is $2-<!--\; -\; -->\frac{{\mathrm{\pi <!--\; \pi \; -->}}^2}{6}$)
Thanks in advance

I'm wondering if someone could help me solve a little conundrum I've been having. I'm trying to find the equation of a line with slope −1 tangent to the curve of a rational function. The function is $y=1/(x-<!--\; -\; -->1)$ and the general equation of the line would be $y=-<!--\; -\; -->x+k$. I understand that if I set the y-values equal -- that is, $1/(x-<!--\; -\; -->1)=-<!--\; -\; -->x+k$ should be able to reorder it into a quadratic and solve for k, giving me the equation of the line. So far, the quadratic I'm finding is $y=-<!--\; -\; -->x^2+x+kx-<!--\; -\; -->k-<!--\; -\; -->1$, yet I can't seem to find a satisfactory value for k that would give me a tangent to the curve. Any thoughts?

How to calculate $\ln <!--\; \; -->(x)$
I know only to calculate $\ln <!--\; \; -->()$ using a calculator, but is there a way to calculate it without calculator:for example: $\ln <!--\; \; -->(4)=??$ as far as I know the only way to do so is to draw the graph of $ln$ but it primarily depends on values from calculator. So really is there some formula for $\ln <!--\; \; -->()$ to calculate its exact value for a specific number?

How to do rational expressions
Never was much of a math student but I am brushing up on my arithmetic and algebra for college.
I am using sample questions from Accuplacer and then using video lectures and practice on Khan Academy. It is going well but sometimes the examples in the lecture are very trivial and I am currently doing very poorly with rational expressions in the form:
$\frac{1}{x+3}+\frac{1}{x}=2x+\frac{3}{x(x+3)}$
or
$\frac{u}{x}+\frac{5u}{x}-<!--\; -\; -->\frac{u}{5x}=\frac{29u}{5x}$
The answers that I supplied are from a multiple choice list, so I need to be able to be able to get those answers on my own.
I have looked up the answers to these problems and I do not understand how it works. I am looking for any killer examples that would really help me grasp it.

A closed form for ${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\ln <!--\; \; -->x\cdot <!--\; \cdot \; -->\ln <!--\; \; -->(1+\frac{1}{2\cosh <!--\; \; -->x})dx$
Is it possible to evaluate this integral in a closed form?
${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\ln <!--\; \; -->x\cdot <!--\; \cdot \; -->\ln <!--\; \; -->(1+\frac{1}{2\cosh <!--\; \; -->x})dx={\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\ln <!--\; \; -->x\cdot <!--\; \cdot \; -->\ln <!--\; \; -->(1+\frac{1}{e^{-<!--\; -\; -->x}+e^x})dx$
I tried to evaluate it with a CAS, and looked up in integral tables, but was not successful.

Logarithms, prove this limit.
Mathematica knows that:
$\log <!--\; \; -->(n)=\underset{s\to <!--\; \to \; -->1}{lim}(1-<!--\; -\; -->\frac{1}{n^{s-<!--\; -\; -->1}})\mathrm{\zeta <!--\; \zeta \; -->}(s)$
Kind of tautological starting with logarithms, but I would like to know better why this limit works:
$\log <!--\; \; -->(n)=\underset{s\to <!--\; \to \; -->1+\frac{2i\mathrm{\pi <!--\; \pi \; -->}k}{\log <!--\; \; -->(n)}}{lim}(1-<!--\; -\; -->\frac{1}{n^{s-<!--\; -\; -->1}})\mathrm{\zeta <!--\; \zeta \; -->}(s-<!--\; -\; -->i\mathrm{\Im <!--\; \Im \; -->}(s))$
for $k$ an integer.
Solving it symbolically for some integer $k$ while leaving $n$ as a variable Mathematica says it is equal to zero. But setting $n$ to any value I get $\log <!--\; \; -->(n)$

Help with logarithmic differentiation problems
$\mathbf{(}\mathbf{1}\mathbf{)}$ Find $y^{\mathrm{\prime <!--\; \prime \; -->}}$ of $y=8^{\sqrt{x}}$
My try:
$\ln <!--\; \; -->y=\ln <!--\; \; -->(8{)}^{\sqrt{x}}$
${\displaystyle \frac{1}{y}}{y}^{\mathrm{\prime <!--\; \prime \; -->}}=\sqrt{x}\ln <!--\; \; -->8$
I don't know how to proceed with right side.
$\mathbf{(}\mathbf{2}\mathbf{)}$ Find $y^{\mathrm{\prime <!--\; \prime \; -->}}$ of $y=(t+4)(t+6)(t+7).$
This one I have no idea what to do so I don't have any work to show. My text says to use logarithmic differentiation, but still I don't how to solve this.
Thank you.

Solving equations with logarithms
I'm having trouble with solving equations that has logarithms in them. For example:
$x^{\log <!--\; \; -->(x)}=\frac{100}{x}$
How can I solve this? I have reed about how to do it but when I try to do the same with this example I just got stuck. The only thing I can think of doing is to draw the lines and see where they cross but I only get one answer $x=10$ but there should be two and I will need to show how I came up with the answer by showing calculations. Please help.

ransform a formula for doubling time
I have a problem with transforming a formula for doubling time. It has two variables: increase (i) and number of periods (p).
The original formula: $p=\frac{\log <!--\; \; -->(2)}{\log <!--\; \; -->(1+i)}$
When I enter $i=0.05$ it correctly calculates $p=14.207$
Now I'm trying to make a formula for "increase". I transformed it to: $i={10}^{\frac{\log <!--\; \; -->(2)}{p}}-<!--\; -\; -->1.$. When I enter $14.207$ periods, it gives me a increase = $0.1189$ (and not 0.05). Where's my mistake?
The transformations I did:
$p=\frac{\log <!--\; \; -->(2)}{\log <!--\; \; -->(1+i)}$
$\log <!--\; \; -->(1+i)=\frac{\log <!--\; \; -->(2)}{p}$
$1+i={10}^{\frac{\log <!--\; \; -->(2)}{p}}$
$i={10}^{\frac{\log <!--\; \; -->(2)}{p}}-<!--\; -\; -->1$
It gives a wrong answer... Do you see where I made a mistake?

Can anyone help me prove the convexity of this rational function? The man who proved the convexity of function used these facts. But I don't know this fact is correct or not. Here are the facts and function:
1. As N increases, f(N) goes to infinity, That implies that there must be a minima (either at N=0 or somewhere else with a finite N)
2. There cannot be more than one (positive) minima since we're dealing with second order equation.
$f(N)=+-<!--\; -\; -->c{N}^4+-<!--\; -\; -->d{N}^3+-<!--\; -\; -->e{N}^2+-<!--\; -\; -->fN+-<!--\; -\; -->g/+-<!--\; -\; -->a{N}^2+-<!--\; -\; -->bN$
a,b,c,d,e,f,g is constants. and $N\ge <!--\; \ge \; -->1$. I guess the second order equation means that between the leading coefficient of the numerator and the denominator is 2.
Are these facts correct? I think fact 1 is no problem, but fact 2 is correct or not.
I am waiting for any answers.

I am trying to calculate the integral
${\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^{\mathrm{\infty <!--\; \infty \; -->}}\frac{a+x}{b^2+(a+x{)}^2}\frac{1}{1+c(a-<!--\; -\; -->x{)}^2}dx$
where $\{a,b,c\}\in <!--\; \in \; -->\mathbb{R}$. I have looked in a table of integrals for rational functions, but with no luck. Is there a smart trick I can utilize?

My analysis book makes the following statement:
Every rational function with real coefficients can be integrated in terms of
1, rational functions,
2. logarithm functions,
3. arctangent functions.
Does it mean that after polynomial division, partial fraction expansion, completing the square, substituting where needed, and knowing that
$\begin{array}{r}\\ \int <!--\; \int \; -->\frac{dx}{(x-<!--\; -\; -->a{)}^n}& =-<!--\; -\; -->\frac{1}{n-<!--\; -\; -->1}\frac{1}{(x-<!--\; -\; -->a{)}^{n-<!--\; -\; -->1}}(n\ne <!--\; \ne \; -->1)\\ \int <!--\; \int \; -->\frac{dx}{x-<!--\; -\; -->a}& =\log <!--\; \; -->|x-<!--\; -\; -->a|\\ \int <!--\; \int \; -->\frac{dx}{x^2+1}& =\arctan <!--\; \; -->x\end{array}$
one can integrate any rational function?

general formula for ${\log }_x<!--\; \; -->(y)$ when $y$ is negative
I'm looking for a general formula for solving a problem of the form ${\log }_x<!--\; \; -->(y)$ when $y<0$. It seems like the formula is $\frac{\ln <!--\; \; -->(|y|)+\mathrm{\pi <!--\; \pi \; -->}i}{\ln <!--\; \; -->(x)}$, but I would like to know how this is derived, or a proof for it. I got this this from looking at the answer of wolfram alpha, so this formula may not be correct.

Solve the inequality $2^{(x^3-<!--\; -\; -->x)}<1$
Let $2^{(x^3-<!--\; -\; -->x)}-<!--\; -\; -->1=f\left(x\right)$
To find the values for which $f(x)<0$ I let $f(x)=0$
$2^{(x^3-<!--\; -\; -->x)}-<!--\; -\; -->1=0$
$2^{(x^3-<!--\; -\; -->x)}=1$
$\ln <!--\; \; -->\left(2^{(x^3-<!--\; -\; -->x)}\right)=\ln <!--\; \; -->\left(1\right)$
$(x^3-<!--\; -\; -->x)\ln <!--\; \; -->\left(2\right)=\ln <!--\; \; -->\left(1\right)$
$(x^3-<!--\; -\; -->x)={\displaystyle \frac{\ln <!--\; \; -->\left(1\right)}{\ln <!--\; \; -->\left(2\right)}}$
$x^3-<!--\; -\; -->x=0$
$x(x^2-<!--\; -\; -->1)=0$
So $f(x)=0$ when $x=0$, $x=1$ and $x=-<!--\; -\; -->1$
After testing values in intervals between those solutions I guess I lucked out and got the answer to be $f(x)<0$ on $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->1)\cup <!--\; \cup \; -->(0,1)$
Although I got the write answer this time I'm not very confident in my ability to do so for other questions, is there a way to do this very "analytically" so you're absolutely sure of the answer and that it holds for all x in the given interval? What tends to be general method? You find the values for which f(x) is zero and take values around that? Are there any tricks you can use? Thanks.

Need help with a proof concerning zero-free holomorphic functions.
Suppose $f(z)$ is holomorphic and zero-free in a simply connected domain, and that $\mathrm{\exists <!--\; \exists \; -->}g(z)$ for which $f(z)=$ exp$(g(z))$
The question I am answering is the following:
Let $t\ne <!--\; \ne \; -->0$ be a complex number. Prove that $\mathrm{\exists <!--\; \exists \; -->}h(z)$ holomorphic such that $f(z)=(h(z){)}^t$
I see that the idea makes sense, but a nudge in the right direction would be appreciated.

Basic demonstration concerning the natural logarithm...
Recently saw a question that asked me to show that $\ln <!--\; \; -->(\sqrt{2}-<!--\; -\; -->1)=-<!--\; -\; -->\ln <!--\; \; -->(\sqrt{2}+1)$. How can I demonstrate that the LHS equals the RHS?