Learn How to Graph Logarithmic Equations

Evaluating Logarithmic Expressions
Evaluate:
$${\log }_4<!--\; \; -->\left({\displaystyle \frac{1}{256}}\right)$$
I am not sure how to approach this since there is nothing set equal to it.

Solve this equation $4^{{\log }_2<!--\; \; -->(x)}-<!--\; -\; -->2^{{\log }_2<!--\; \; -->(x)}=3^{{\log }_3<!--\; \; -->(12)}$
I thought to write $2^{{\log }_2<!--\; \; -->(x{)}^2}-<!--\; -\; -->2^{{\log }_2<!--\; \; -->(x)}=3^{{\log }_3<!--\; \; -->(12)}$. Then is there a way to factorize $2^{{\log }_2<!--\; \; -->(x)}$? I don't know how to proceed...

Quick Log problem?
$2^{(x^3)}=3^{(x^2)}$
Solve for x
I'm pretty sure I use logs to solve this, but how? to what base? I'm kinda lost.. Thanks

Solving a second-degree exponential equation with logarithms
The following equation is given:
$8^{2x}+8^x-<!--\; -\; -->20=0$
The objective is to solve for x in terms of the natural logarithm ln.
I approach as follows:
${\log }_8<!--\; \; -->(8^{2x})={\log }_8<!--\; \; -->(-<!--\; -\; -->8^x+20)$
$2x={\log }_8<!--\; \; -->(-<!--\; -\; -->8^x+20)$
$2x={\displaystyle \frac{\ln <!--\; \; -->(-<!--\; -\; -->8^x+20)}{\ln <!--\; \; -->8}}$
$x={\displaystyle \frac{\ln <!--\; \; -->(-<!--\; -\; -->8^x+20)}{2\ln <!--\; \; -->8}}$
and at this point I'm unable to proceed.

Find distribution mean from the mean and sd of the log
I have a distribution with a long tail and use a model to predict the mean and standard deviation of its log.
Given the mean and standard deviation of the log, how do I find the mean of the actual distribution?

Logarithmic Equation: How to solve for x
Equation:
$${\log }_a<!--\; \; -->(x)+{\log }_a<!--\; \; -->(x-<!--\; -\; -->4)={\log }_a<!--\; \; -->(x+6)$$
Progress
$${\log }_a<!--\; \; -->(x^2-<!--\; -\; -->4x)={\log }_a<!--\; \; -->(x+6)$$
$$x^2-<!--\; -\; -->5x-<!--\; -\; -->6=0$$
Delta
$$x1=6$$
$$x2=-<!--\; -\; -->1$$

Why is $\ln <!--\; \; -->(x^x)=x\ln <!--\; \; -->(x)$ valid?
I know that $\ln <!--\; \; -->(x^k)=k\ln <!--\; \; -->(x)$ for any constant $k$, but why is $\ln <!--\; \; -->(x^x)=x\ln <!--\; \; -->(x)$. The exponent $x$ is not constant.

Can a logarithm have a function as a base?
For example is ${\log }_{\sin <!--\; \; -->(x)}<!--\; \; -->(3x)$ a ridiculous equation?
I couldn't find an example on any page about logarithms that used a function on a base, but it seems that for an equation like $\sin <!--\; \; -->(x{)}^{12x}$, the log's base would have to be the sine function. Thank you for the advice!

Differentiating this problem $\frac{2t^{3/2}}{\ln <!--\; \; -->(2t^{3/2}+1)}$
How does one differentiate the function
$$y(t)=\frac{2t^{3/2}}{\ln <!--\; \; -->(2t^{3/2}+1)}.$$
How do I start/process solving this? Do i take the ln of both side? If so I get the log of the top - the log of the bottom. which is the log of a log? If I do the quotient rule right away, i get the log expression in the bottom squared. Help please?

Need to simplify a logarithmic expression
Can someone simplify this ($\log$ here refers to the common logarithm)?
$\sqrt{4\log <!--\; \; -->2+(\log <!--\; \; -->5{)}^2}+\sqrt{4\log <!--\; \; -->5+(\log <!--\; \; -->2{)}^2}$
I know this has a simple solution but I cannot find it.

Evaluate. $8^{3x}=360$

Why does $n^{\ln <!--\; \; -->\ln <!--\; \; -->n}=(\ln <!--\; \; -->n{)}^{\ln <!--\; \; -->n}$?
Note: this is (part of my solution to) a homework question. Please DO NOT tell me the answer!
I am trying to compare the following functions:
$$n^{\ln <!--\; \; -->\ln <!--\; \; -->n}(\ln <!--\; \; -->n{)}^{\ln <!--\; \; -->n}$$
It appears that they are equal (assuming $n>1$), but I have absolutely no idea why this would be the case.
I am missing something really obvious and I have been hitting my head on this for about 45 minutes. A hint would be appreciated.

Inverse of logarithmic function
I have this problem: $f:(0,\mathrm{\infty <!--\; \infty \; -->})\to <!--\; \to \; -->\mathbb{R}$
$$f(x)=x+1+\ln <!--\; \; -->x$$
g is the reverse of f.
I tried to solve the problem like this $y=f(x);x+\ln <!--\; \; -->x=y-<!--\; -\; -->1$ and I got stuck
The problem is:
$g(2)=$? Also g'(2) =? Anyone got any idea?

Meaning of exponent in logarithm?
I have this particular difficulty :
$${\log }_b^a<!--\; \; -->(c)=x$$
I know it is different from power of base ${\log }_{b^a}<!--\; \; -->(c)=x$, but what does it actually mean?
The actual question that i got in paper was
Find value of
$$\sqrt{{\log }_{0.5}^2<!--\; \; -->8}$$
And its answer was given as 3.

Condense the expression to the logarithm of a single quanity.
$${\log }_7<!--\; \; -->(5x)-<!--\; -\; -->8{\log }_7<!--\; \; -->(x)$$

Best way to handle the ratio which cannot be represented as floating point numbers.
I need to calculate the ratio of the form:
$s=\underset{1}{\overset{3}{\sum <!--\; \sum \; -->}}{q}_i$, $$$p_i=\frac{q_i}{\underset{1}{\overset{3}{\sum <!--\; \sum \; -->}}{q}_i}$
where $q_i>0$. One problem is that $q_i$ are too small that they can not represented as floating point numbers, then I can try logarithms $z_i=\mathrm{l}\mathrm{o}\mathrm{g}(q_i)$ and $\mathrm{l}\mathrm{o}\mathrm{g}(p_i)=\mathrm{l}\mathrm{o}\mathrm{g}(q_i)-<!--\; -\; -->\mathrm{l}\mathrm{o}\mathrm{g}(q_1+q_2+q_3)$, at this moment, I know $\mathrm{l}\mathrm{o}\mathrm{g}(q_1)=-<!--\; -\; -->2012,\mathrm{l}\mathrm{o}\mathrm{g}(q_2)=-<!--\; -\; -->2013,\mathrm{l}\mathrm{o}\mathrm{g}(q_3)=-<!--\; -\; -->2014$, but how to deal with $\mathrm{l}\mathrm{o}\mathrm{g}(q_1+q_2+q_3)$, any one could give me some idea?
thanks

How to evaluate square of logarithms to solve s(n) = O(a(n))?
I've never used log before, nor worked with big-O notation, so I'm completely useless at this stuff. Any, any, any help or direction you can give would be helpful as the professor hasn't covered this in the least and this is the type of thing he's throwing at us on exams:
Find a(n) where s(n) = O(a(n)) for:
$s(n)=(n{\log }_2<!--\; \; -->n+1{)}^2$
I just have no idea where to even begin.
By way of apology / context, I've read my discrete math text book chapter on log and big-O notation 8-10 times now, scoured Google for a full day, banged my head against this for hours more and come up with nothing. I don't know how but I'm in my 4th year of college and somehow none of my classes covered logarithms and I'm just not finding good resources on how to learn this stuff, so here I am for help. (I've given up on passing the class already, just trying to keep learning so I can hopefully pass when I retake next semester.)

Find the product of positive roots of equation $\sqrt{2008}{x}^{{\log }_{2008}<!--\; \; -->x}=x^2$
Problem : Find the product of positive roots of equation $\sqrt{2008}{x}^{{\log }_{2008}<!--\; \; -->x}=x^2$
Solution :
The given equation can be written as $\sqrt{2008}{x}^{{\log }_{2008}<!--\; \; -->x}=x^2$
$⟹<!--\; ⟹\; -->\sqrt{2008}{x}^{{\log }_x<!--\; \; -->{2008}^{-<!--\; -\; -->1}}=x^2$
$⟹<!--\; ⟹\; -->{\displaystyle -<!--\; -\; -->\sqrt{2008}\frac{1}{2008}=x^2}$ [by using $a^{{\log }_a<!--\; \; -->m}=m$]
$⟹<!--\; ⟹\; -->{\displaystyle x^2=-<!--\; -\; -->\frac{1}{\sqrt{2008}}}$
Now how to find the product of positive roots please guide, thanks.

Series for logarithms
This is more of a challenge than a question, but I thought I'd share anyway. Prove the following identities, and prove that the pattern continues.
$\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}(\frac{1}{2n+1}-<!--\; -\; -->\frac{1}{2n+2})=\ln <!--\; \; -->2$
$\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}(\frac{1}{3n+1}+\frac{1}{3n+2}-<!--\; -\; -->\frac{2}{3n+3})=\ln <!--\; \; -->3$
$\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}(\frac{1}{4n+1}+\frac{1}{4n+2}+\frac{1}{4n+3}-<!--\; -\; -->\frac{3}{4n+4})=\ln <!--\; \; -->4$
$\mathrm{e}\mathrm{t}\mathrm{c}.$

I need help on the process of solving this derivative.
How do I go about solving this derivative.
$f(x)=\ln <!--\; \; -->\left(\frac{7x}{x+4}\right)$
I go from this to
$1.f(x)=\ln <!--\; \; -->(7)+\ln <!--\; \; -->(x)-<!--\; -\; -->\ln <!--\; \; -->(x+4)$
and then
$2.f^{\prime }(x)=\frac{1}{x}-<!--\; -\; -->\frac{1}{x+4}$
then
$3.f^{\prime }(x)=\frac{x+4-<!--\; -\; -->x}{x(x+4)}$
and end up with
$4.f^{\prime }(x)=\frac{4}{x(x+4)}$
which is the correct answer. This procedure is what we did in class, but I do not get how I went from step 2 to 3 which is what has me confused. A step by step explanation would be greatly appreciated. Thanks!