Learn How to Graph Logarithmic Equations

How to simplify $3^{(2{\log }_3<!--\; \; -->35)}$
How do I simplify this? This is what I have done so far:
$2{\log }_3<!--\; \; -->5={\log }_3<!--\; \; -->5^2={\log }_3<!--\; \; -->(25)$
$3^{{\log }_3<!--\; \; -->(25)}$
What do I do from here? And the answer is one of these mixed solutions:
$0$
$-<!--\; -\; -->2$
$-<!--\; -\; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}$
$\frac{1}{x+2}$
$\pm <!--\; \pm \; -->\frac{4}{25}$
$25$
$30°$
$2$
$3$
$5$
$\mathrm{\pi <!--\; \pi \; -->}$
$\frac{\mathrm{\pi <!--\; \pi \; -->}}{3}$
$(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},2)$
$4(x+1{)}^2+3$
$-<!--\; -\; -->\frac{\sqrt{2}}{2}$
$-<!--\; -\; -->\frac{\sqrt{3}}{2}$
$\frac{\sqrt{2}}{2}$

Why does ${\int <!--\; \int \; -->}_a^{ab}\frac{1}{x}dx={\int <!--\; \int \; -->}_1^b\frac{1}{t}dt$?
I can't understand how the integral having limits from $a$ to $ab$ in Step 1 is equivalent to the integral having limits from $1$ to $b$. I'm a beginner here. Please explain in detail.
$\begin{array}{rl}\ln <!--\; \; -->(ab)={\int <!--\; \int \; -->}_1^{ab}\frac{1}{x}dx& ={\int <!--\; \int \; -->}_1^a\frac{1}{x}dx+{\int <!--\; \int \; -->}_a^{ab}\frac{1}{x}dx\\ & ={\int <!--\; \int \; -->}_1^a\frac{1}{x}dx+{\int <!--\; \int \; -->}_1^b\frac{1}{at}d(at)\\ & ={\int <!--\; \int \; -->}_1^a\frac{1}{x}dx+{\int <!--\; \int \; -->}_1^b\frac{1}{t}dt\\ & =\ln <!--\; \; -->(a)+\ln <!--\; \; -->(b).\end{array}$

Question releating to the ${\int <!--\; \int \; -->}_1^x\frac{\ln <!--\; \; -->(t)}{t+1}$
If $f(x)={\int <!--\; \int \; -->}_1^x\frac{\ln <!--\; \; -->(t)}{t+1}dt$ if $x>0$. Compute $f(x)+f(1/x)$. As a check, you should obtain $f(2)+f(1/2)=(\ln <!--\; \; -->2{)}^2$
I have tried evaluating the integral ${\int <!--\; \int \; -->}_1^x\frac{\ln <!--\; \; -->(t)}{t+1}dt$ by parts, but it turns out to be futile as $\int <!--\; \int \; -->\frac{\ln <!--\; \; -->(t)}{t+1}dt=\ln <!--\; \; -->(t)\ln <!--\; \; -->(t+1)-<!--\; -\; -->\int <!--\; \int \; -->\frac{\ln <!--\; \; -->(t+1)}{t}t$ and $\int <!--\; \int \; -->\frac{\ln <!--\; \; -->(t+1)}{t}=\ln <!--\; \; -->(t)\ln <!--\; \; -->(t+1)-<!--\; -\; -->\int <!--\; \int \; -->\frac{\ln <!--\; \; -->(t)}{t+1}dt$ which leads to $\int <!--\; \int \; -->\frac{\ln <!--\; \; -->(t)}{t+1}dt=\int <!--\; \int \; -->\frac{\ln <!--\; \; -->(t)}{t+1}dt$
Is there any other way to evaluate this integral?

Solving $e^{4x}+3e^{2x}-<!--\; -\; -->28=0$
How to solve this equation:
$e^{4x}+3e^{2x}-<!--\; -\; -->28=0$
I don't know how to solve this problem. I read over another example, $e^{2x}-<!--\; -\; -->2e^x-<!--\; -\; -->8=0,$ and it said that $e^{2x}$ is $e$ to the $x$ squared, so $e^{2x}=e^{x^2}$. Why is this so? So $e^{4x}=e^{x^4}$?
Please help me to solve it.

Logorithms on a first level learning
Solve log${}_{5x-<!--\; -\; -->1}$ $4$$=$$1/3$
$(5x-<!--\; -\; -->1{)}^{1/3}$=4
$((5x-<!--\; -\; -->1{)}^{1/3}{)}^3$ = $4^3$
$5x-<!--\; -\; -->1=64$
$5x=65$
$13$
I am not sure where to go with this. I learned some things about logs before my class ended for the year. I just wanted to expand on my knowledge. This is more advance than what I am use to doing. Can someone please show me.

Subtracting a constant from log-concave function preserves log-concavity, if the difference is positive
I am trying to work out a question from 'Convex Optimization - Boyd' . Specifically, exercise 3.48:
Show that if $f:{\mathbb{R}}^n\to <!--\; \to \; -->\mathbb{R}$ is log-concave and $a>0$, then the function $g=f-<!--\; -\; -->a$ is log-concave, where $\mathrm{dom}<!--\; \; -->g=\{x\in <!--\; \in \; -->\mathrm{dom}<!--\; \; -->f|f(x)>a\}?$
This is my attempt so far:
By assumption, we have
$\frac{d^2\log <!--\; \; -->f}{d{x}^2}=\frac{f^{″}f-<!--\; -\; -->(f^{\prime }{)}^2}{f^2}\le <!--\; \le \; -->0.$
Considering $g=f-<!--\; -\; -->a$, we have
$\frac{d^2\log <!--\; \; -->g}{d{x}^2}=\frac{f^{″}(f-<!--\; -\; -->a)-<!--\; -\; -->(f^{\prime }{)}^2}{(f-<!--\; -\; -->a{)}^2}=\frac{f^{″}f-<!--\; -\; -->(f^{\prime }{)}^2}{(f-<!--\; -\; -->a{)}^2}-<!--\; -\; -->\frac{f^{″}a}{(f-<!--\; -\; -->a{)}^2}.$
If $f^{″}>0$, then we are complete. I'm not sure how to handle the alternative situation though, can anyone help?

Upper bound of natural logarithm
I was playing looking for a good upper bound of natural logarithm and I found that
$\ln <!--\; \; -->x\le <!--\; \le \; -->x^{1/e}$
apparently works: Can someone give me a formal proof of this inequality?

Evaluate ${\int <!--\; \int \; -->}_1^{\mathrm{\infty <!--\; \infty \; -->}}\frac{\ln <!--\; \; -->(2x-<!--\; -\; -->1)}{x^2}$
My approach is to calc
${\int <!--\; \int \; -->}_1^X\frac{\ln <!--\; \; -->(2x-<!--\; -\; -->1)}{x^2}dx$
and then take the limit for the answer when $X\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$
However, I must do something wrong. The correct answer should be $2\ln <!--\; \; -->(2)$
${\int <!--\; \int \; -->}_1^X\frac{\ln <!--\; \; -->(2x-<!--\; -\; -->1)}{x^2}dx={[-<!--\; -\; -->\frac{1}{x}\ln <!--\; \; -->(2x-<!--\; -\; -->1)]}_1^X+{\int <!--\; \int \; -->}_1^X\frac{1}{x}\times <!--\; \times \; -->\frac{2}{2x-<!--\; -\; -->1}dx=-<!--\; -\; -->\frac{1}{X}\ln <!--\; \; -->(2X-<!--\; -\; -->1)+2{\int <!--\; \int \; -->}_1^X-<!--\; -\; -->\frac{1}{x}+\frac{2}{2x-<!--\; -\; -->1}dx=-<!--\; -\; -->1\frac{1}{X}\ln <!--\; \; -->(2X-<!--\; -\; -->1)-<!--\; -\; -->2\ln <!--\; \; -->X+2\ln <!--\; \; -->(2X-<!--\; -\; -->1)$
Am I wrong? If I'm not, how to proceed?
=== EDIT ===
After the edit I wonder if this is the correct way to proceed:
$-<!--\; -\; -->\frac{1}{X}\ln <!--\; \; -->(2X-<!--\; -\; -->1)-<!--\; -\; -->2\ln <!--\; \; -->X+2\ln <!--\; \; -->(2X-<!--\; -\; -->1)$
The first part will do to zero because of $\frac{1}{X}$ so we ignore that one, the second and third part:
$-<!--\; -\; -->2\ln <!--\; \; -->X+2\ln <!--\; \; -->(2X-<!--\; -\; -->1)=2\ln <!--\; \; -->\left(\frac{2X-<!--\; -\; -->1}{X}\right)=2\ln <!--\; \; -->(2-<!--\; -\; -->\frac{1}{X})=2\ln <!--\; \; -->(2)$

Does loga/logb = log(a^(1/logb))?
I know $\log <!--\; \; -->(a^b)=b\log <!--\; \; -->(a)$
However, Wolfram Alpha tells me that $\frac{\log <!--\; \; -->(a)}{\log <!--\; \; -->(b)}$ does not equal $\log <!--\; \; -->(a^{\frac{1}{\log <!--\; \; -->(b)}})$
Is Wolfram Alpha correct? If it is, why is it correct?
I'm using base 10 logs instead of natural logs, although I doubt it makes a difference.

Homework help to rearrange formula
Given the equation
$V_m=u(\ln <!--\; \; -->m_0-<!--\; -\; -->\ln <!--\; \; -->m_8)-<!--\; -\; -->g{t}_f$
I need to solve for $m_0$ Here is what I have but it looks messy and I feel like there is sometihng wrong or a better way
1st attempt
$\begin{array}{rl}& V_m=u(\ln <!--\; \; -->m_0-<!--\; -\; -->\ln <!--\; \; -->m_8)-<!--\; -\; -->g{t}_f\\ & V_m=u(\ln <!--\; \; -->m_0)-<!--\; -\; -->u(\ln <!--\; \; -->m_8)-<!--\; -\; -->g{t}_f\\ & V_m=u(\ln <!--\; \; -->m_0)-<!--\; -\; -->u(\ln <!--\; \; -->m_8)-<!--\; -\; -->g{t}_f\\ & V_m=u(\ln <!--\; \; -->m_0)-<!--\; -\; -->u(\ln <!--\; \; -->m_8)-<!--\; -\; -->g{t}_f\\ & u(\ln <!--\; \; -->m_0)-<!--\; -\; -->u(\ln <!--\; \; -->m_8)-<!--\; -\; -->g{t}_f-<!--\; -\; -->V_m=0\\ & u(\ln <!--\; \; -->m_0)-<!--\; -\; -->u(\ln <!--\; \; -->m_8)-<!--\; -\; -->V_m=g{t}_f\\ & u(\ln <!--\; \; -->m_0)=g{t}_f+u(\ln <!--\; \; -->m_8)+V_m\\ & \ln <!--\; \; -->m_0=(g{t}_f+u(\ln <!--\; \; -->m_8)+V_m)÷<!--\; ÷\; -->u\\ & e^{(g{t}_f+u(\ln <!--\; \; -->m_8)+V_m)÷<!--\; ÷\; -->u}=m_0\end{array}$
2nd attempt - think this looks a little better but still not there yet
$\begin{array}{rl}& V_m=u(\ln <!--\; \; -->\frac{m_0}{m_8})-<!--\; -\; -->g{t}_f\\ & V_m+g{t}_f=u(\ln <!--\; \; -->\frac{m_0}{m_8})\\ & \frac{V_m+g{t}_f}{u}=\ln <!--\; \; -->\frac{m_0}{m_8}\\ & e^{\frac{V_m+g{t}_f}{u}}=\frac{m_0}{m_8}\\ & m_8{e}^{\frac{V_m+g{t}_f}{u}}=m_0\end{array}$

For which $z$ is the following true: $Log(i{z}^2)=\frac{i\mathrm{\pi <!--\; \pi \; -->}}{2}+2Log(z)$. We raised both sides by $e$ and concluded this equation holds for all $z\ne <!--\; \ne \; -->0$, but are not sure if we have all the work for it. We are mostly concerned with how the Arg part comes into play here and if that will limit what $z$ can be.

Pre-calculus algebra logarithm question
I don't understand how to solve this equation. Been struggling with it and don't know how to start:
${\log }_2<!--\; \; -->x=8+9{\log }_x<!--\; \; -->2$
Can someone please help me out?

solve the equation using logarithms (I think this is easy level)
Solve the equation for $x$ by using base 10 logarithms.
$16\cdot <!--\; \cdot \; -->4^{2.5x}=9$
EDIT: I made a typo (somehow... I was very far off!!)
The correct equation is this:
$16\cdot <!--\; \cdot \; -->4^{2.5x}=70$
Can it be written like:
$2.5x{\log }_{10}<!--\; \; -->(5)=70?$
Then get:
${\log }_{10}<!--\; \; -->(5)=\frac{70}{2.5x}$
The computer wants a largest value and smallest value, similar to an answer for a quadratic problem. I need to know how to get the 2 answers even if one ends up negative (I know the negative will be tossed out, but I still need to know how to get the answer).

Is $\ln <!--\; \; -->\sqrt{2}$ irrational?
I know that the natural log of any positive algebraic number is transcendental, as a consequence of the Lindemann-Weierstrass theorem, but what about the natural log of the square root of two (which is irrational).
Is this rational or irrational?

It's on Indefinite Integrals
$\int <!--\; \int \; -->\sqrt{1+2\tan <!--\; \; -->x(\tan <!--\; \; -->x+\sec <!--\; \; -->x)}dx$
Please tell me the way of solving such questions. like what could i assume sec x or sec x tan x to be equal to?

Calculate ln(x) using 8-digit calculator
I have a bit of a unique problem. Well, maybe not a problem because I'm really just curious about it, but...
I have a simple 8 digit calculator. It has +, -, x, /, and a constant operation function. It has no usable memory slot.
I need (lit. would like to know how) to calculate ln(x) on this calculator. Any series I have found lose a lot of precision since whatever is outside the 8 digit range is truncated without rounding. I have used Padé approximants with trig functions, and a continued fraction for exp(), but none of these two methods are adequate for ln(x), either because there is no way around the truncation error or it requires writing down many intermediate values, or it requires memorizing and punching in many constants.
Does anyone know any good way to calculate ln(x) (or any other base, since memorizing one constant is OK) to as many digits as possible as easily as possible on this calculator (bonus points to methods where the value on the screen is immediately, meaning you don't need to write it down)?
For that matter, if you happen to know some good ways to calculate other special functions, I am interested in that as well.
Thanks!

Using the natural logarithm, solve for x in the equation $$3^x=9$$

How do I simplify $\log <!--\; \; -->(1/\sqrt{1000})$?
What I have done so far:
1) Used the difference property of logarithms
$\log <!--\; \; -->\left({\displaystyle \frac{1}{\sqrt{1000}}}\right)=\log <!--\; \; -->(1)-<!--\; -\; -->\log <!--\; \; -->(\sqrt{1000})$
2) Used the exponent rule for logarithm
$\log <!--\; \; -->(1)-<!--\; -\; -->\frac{1}{2}\log <!--\; \; -->(1000)$
I'm stuck at this point. Can someone explain why and what I must do to solve this equation?

Write as the sum/difference of logarithm
$${\log }_b<!--\; \; -->(\frac{A^3}{L_P})$$

Write in logarithmic form :
$$3^4=81$$