Learn How to Graph Logarithmic Equations

$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(\frac{\ln <!--\; \; -->(x-<!--\; -\; -->2)}{\ln <!--\; \; -->(x-<!--\; -\; -->1)}{)}^{x\ln <!--\; \; -->x}.$
L'Hopital seems like a very hardcore solutions given the situation.Are the any other options?

Can anyone help me in proving this?
As we know $e^{\ln <!--\; \; -->(x)}=x$
To prove this I applied $\ln$ on LHS (left hand side) and I got RHS (rights hand side ) as follows
$\ln <!--\; \; -->(e^{\ln <!--\; \; -->(x)})=\ln <!--\; \; -->(x)$
$\ln <!--\; \; -->(x)\ln <!--\; \; -->(e)=\ln <!--\; \; -->(x)$
$\ln <!--\; \; -->(x)=\ln <!--\; \; -->(x)$
But I didn't knew how the first step or the equation which is to be proved. Can anyone help me in this.

Is $1+\lg <!--\; \; -->i=\lg <!--\; \; -->(i+i)$?
I've been studying Sedgewick's "Algorithms" book and in proof of one proposition he writes the following:
the property is preserved because
$1+\lg <!--\; \; -->i=\lg <!--\; \; -->(i+i)\le <!--\; \le \; -->\lg <!--\; \; -->(i+j)=\lg <!--\; \; -->k$
I cannot wrap my brain around the first part of this inequation, namely $1+\lg <!--\; \; -->i=\lg <!--\; \; -->(i+i)$. Can anyone offer an explanation? Thanks in advance!

Prove that ${\log }_a<!--\; \; -->(1/x)=-<!--\; -\; -->{\log }_a<!--\; \; -->(x)$
But it has two problems: when x=0 and on the other problem it doesn't mention any condition. How should I solve it in each of them?

To find, wether '1' lies in the range of f, where $f(x)=[ln(\frac{7x-<!--\; -\; -->x^2}{12}){]}^{\frac{3}{2}}$?
For the given function, the question is whether, f(x) can equal 1 for some real value of x?

Numbers of solutions of the equation ${\log }_3<!--\; \; -->\frac{2x^2+3x+3}{5}=\frac{1}{{\log }_{2x^2+3x+9}<!--\; \; -->9}$
Pretty straightforward question. When I solved it, I got two positive and two negative solutions, so that would make 4 in total. None get discarded as the arguments in the logarithm still stay positive.
However, the solution is supposed to be 2, so I'm not sure where I went wrong.
I got the values of $x$ as $-<!--\; -\; -->2$, $1/2$, $12/5$ and $-<!--\; -\; -->18/5$

A definite integral contianing ln(x)
everyone, I met a tough definite integral as follows,
$I=\underset{1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\int <!--\; \int \; -->}}\frac{\ln <!--\; \; -->x}{{\left(x+a\right)}^m{\left(x+b\right)}^{n+1}}dx,$
where a and b are constant, m and n are non-negative integers.

Inverting the exponential decay equation?
How would I algebraically isolate $x$ in the following exponential decay equation, so that $x$ is most easily derived?
$y=A{e}^{-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}x}$

Solving ${\log }_2<!--\; \; -->(x^2)=x$ explicitly?
I'm having problems getting a proper step-by-step solution to this equation.
${\log }_2<!--\; \; -->(x^2)=x$
I know the results are 2 and 4, but so far I can get only solutions like these:
$2^x=x^2\text{or}\frac{\log <!--\; \; -->(x)}{x}=\frac{\log <!--\; \; -->(2)}{2}.$
I'd like something along the lines of $x=\cdots <!--\; \cdots \; -->$
Is it possible with some basic knowledge?

Derivative of log and exponential function
How would you find the derivative of:
$y=e^{\sqrt{x}}+\ln <!--\; \; -->\sqrt{x}$
Would I simply use the product rule?

Algebra: logarithms word question
Initially there are $2000$ bacteria in a given culture. The number of bacteria $N$ is tripling every hour so $N=2000\cdot <!--\; \cdot \; -->3^t$, where $t$ is the measurable in hours.
a) How many bacteria are present after 4 hours?
b) How long is it until there are 1000000 bacteria?

What $n^{\frac{1}{{\log }_2<!--\; \; -->n}}$ means?
I was confused with about the $n^{\frac{1}{{\log }_2<!--\; \; -->n}}$ expression. I am not sure how to make mathematical sense of it - i.e. express it in another way for easier understanding. I tried to plug in some numbers like 4 and 8. I got the answer 2 in both cases. Is it what this expression means? 2?

Ambiguity with Logs and Inequalities
Solve the following inequality:
${0.8}^x>0.4$
Method 1 (Using the Common Logarithm)
${\log }_{10}<!--\; \; -->{0.8}^x>{\log }_{10}<!--\; \; -->0.4$
$x{\log }_{10}<!--\; \; -->0.8>{\log }_{10}<!--\; \; -->0.4$
Because
${\log }_{10}<!--\; \; -->0.8<0$
The sign of the inequality will change when we divide both sides of the equation by the LHS of the above inequality:
$x<\frac{{\log }_{10}<!--\; \; -->0.4}{{\log }_{10}<!--\; \; -->0.8}$
Using a reverse of the change of base theorem, we get:
$x<{\log }_{0.8}<!--\; \; -->0.4$
Method 2 (Using another based Logarithm)
${\log }_{0.8}<!--\; \; -->{0.8}^x>{\log }_{0.8}<!--\; \; -->0.4$
$x{\log }_{0.8}<!--\; \; -->0.8>{\log }_{0.8}<!--\; \; -->0.4$
Because
${\log }_{0.8}<!--\; \; -->0.8>0$
The sign of the inequality will not change when we divide both sides of the equation by the LHS of the above inequality:
$x>\frac{{\log }_{0.8}<!--\; \; -->0.4}{{\log }_{0.8}<!--\; \; -->0.8}$
The denominator of the RHS of the inequality is equal to one, therefore:
$x>{\log }_{0.8}<!--\; \; -->0.4$
The sign is reversed using both methods - the textbook states the answer to be the one gotten using the first method, but I can't spot where I went wrong in method two.
Any help will be greatly appreciated, thanks in advance.

Solve ${\log }_2<!--\; \; -->x={\log }_4<!--\; \; -->(x+6)$ for $x$ using the change of base formula.
I already tried changing the base on both sides but that didn't work I know it must be in the form of a quadratic for a substitution to be made.

What is the largest $n$-digit number which is also an exact $n$th power?
For example,the largest $2$-digit number which is an exact $2$nd power is $81$

Show that $g(x)=x\ln <!--\; \; -->x$ and $g(x)=e^x$ are bounded below.
Show that $g(x)$ is bounded below, for $0\le <!--\; \le \; -->x$
a) $g(x)=\{\begin{array}{ll}0& \text{if }x=0\\ x\ln <!--\; \; -->x& \text{if }x>0\end{array}$
b) $g(x)=e^x$
For (a), $g(x)\ge <!--\; \ge \; -->0$ for $x\ge <!--\; \ge \; -->0$
For (b), $g(x)\ge <!--\; \ge \; -->1$ for $x\ge <!--\; \ge \; -->0$
Is that all I have to say? Or is there a more technical definition/proof?

Proving a logarithmic inequality
I'm interested why this is true:
$\text{Considering }\mathrm{\forall <!--\; \forall \; -->}(x,y,z)\in <!--\; \in \; -->(1,\mathrm{\infty <!--\; \infty \; -->})$
The following holds:
${\log }_x<!--\; \; -->y^z+{\log }_x<!--\; \; -->z^y+lo{g}_y{z}^x\ge <!--\; \ge \; -->\frac{3}{2}$
This is taken from a high school textbook of mine. I tried finding a meaningful manipulation by using AM-GM, but that got pretty messy. I'd like to avoid Lagrange multipliers since this is meant to be a pretty basic problem.
Any progress would be appreciated.

Find the leftmost (most significant digits) of a large exponent calculation, say ${99}^{99}$
I want to find the initial 10 digits of an exponent calculation whose result is a very large number - Say, ${99}^{99}=3.697296\times <!--\; \times \; -->{10}^{197}$
I only need to know the digits $3697296$
Is there any way of finding this without doing the full calculation?

Logarithm rule for branch cut logarithms
I know that for $a,b\in <!--\; \in \; -->\mathbb{R}$, the rule $\log <!--\; \; -->(ab)=\log <!--\; \; -->(a)+\log <!--\; \; -->(b)$
What about for $a_1,b_1$ in the right half-plane, or $a_2,b_2$ in the sector from $\frac{-<!--\; -\; -->3\mathrm{\pi <!--\; \pi \; -->}}{4}$ to $\frac{3\mathrm{\pi <!--\; \pi \; -->}}{4}$? For the logarithm, take a branch cut on the negative real line.

How to solve logarithm word problem given the exponential equation?
The question is
The world population in 2000 was approximately 6.08 billion. The annual rate of increase was about 1.26%. The function that models this is
$y=6.08(10{)}^{.0052t}$
where y represents the population in billions and t is the time in years after 2000. Write and solve a logarithmic equation to determine what year the world population will exceed 7.5 billion.
So is the question asking me to convert the given equation into logarithmic equation than
$7.05=6.08(10{)}^{.0052t}$
( 7.5 billion is y right?)
$7.05/6.08={10}^{.0052t}$
$1.15={10}^{.0052t}$
$log(1.15)=.0052t$
How would i even solve after that? is the way I've done so far good? how would i proceed after this than?