Learn How to Graph Logarithmic Equations

Diffie hellman and the discrete algorithm problem
Suppose Alice and Bob are exchanging keys using Diffie-Hellman Key-Exchange Algorithm.
a - Alice secret key
g - generator
p - prime
x - the public key passed from Alice to Bob.
Eve is listening to the communication and she is exposed to the three parameters g,p,x.
She's using a brute-force method to find $a$, Alice's secret key.
Thus, Eve is looking for a satisfying this equation:
$g^{a}modp=x$
Now, I know (By testing) Eve can find $a^{\prime }\ne <!--\; \ne \; -->a$ satisfying the equation above, and by using $a^{\prime }$ she can also compute the common secret key used by Alice and Bob.
Why is it mathematically true?

Next number that is a 1 and zeroes
There's probably a much better way of expressing this, but I don't know it, so I guess that's part of the question too. I'm not even sure what to tag it.
How do I find the next number greater than a given number that is a 1 followed by zeroes.
For instance
12 => 100
9343 => 10,000
etc.
Thanks

find the domain of root of a logarithmic function
I'm a little confused about this question since output of a logarithmic function varies from $-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}$ to $\mathrm{\infty <!--\; \infty \; -->}$ .I should find the domain of this function: $y=\sqrt{{\log }_x<!--\; \; -->(10-<!--\; -\; -->x^2)}$ . How can I find the interval that makes ${\log }_x<!--\; \; -->(10-<!--\; -\; -->x^2)$ greater than zero?

${\log }_{-<!--\; -\; -->3}<!--\; \; -->\left(9\right)$
Can you have a negative base for a logarithm? If the answer is no, why can you not? Isn't the answer to the above just $2$?
Similarly, is $-<!--\; -\; -->3$ also a solution to the equation ${\log }_x<!--\; \; -->\left(9\right)=2$ (I know one solution is obviously 3).

Please explain this simple rule of logarithms to me
Right I know this one is simple and I know that I just need a push to make it sink in in my head..
I am studying control systems and in one of the tutorial examples the tutor says
Show that
$20\log <!--\; \; -->(1/x)=-<!--\; -\; -->20\log <!--\; \; -->(x)$
I know that when you have a divide or a multiply with logarithms you add them and subtract them but for my own understanding I just need someone to like slowly show me how this works..
If I take the log of the numerator I have $20\log <!--\; \; -->(1)=0$ but I don't know where to go from here.. So do I now just take the log of the denominator and as the numerator was zero it is just minus whatever the log of the denominator is... Getting myself a bit muddled.. Thanks

$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}n\sqrt{2}(\sqrt{\ln <!--\; \; -->(n+1)}-<!--\; -\; -->\sqrt{\ln <!--\; \; -->n})=0$
But I haven't any ideas how to do it... My calculations shows that this sequence is monotonously decreasing.
I've proved that using inequality $\ln <!--\; \; -->(1+x^a)\le <!--\; \le \; -->ax$ and double-sided theorem.

Rewrite a formula in terms of exponential to the power of logarithm
I would like to rewrite the following formula, f(x).
how can I rewrite the f(x)
$f(x)=\frac{(1+2B)}{2A\sqrt{B^2+(1+2B)\frac{x}{A}}}$
as
${\exp }^{\log <!--\; \; -->(f(x)}$

Solving this logarithm equation?
How do I solve this equation using common logarithms?
$\log <!--\; \; -->x=1-<!--\; -\; -->\log <!--\; \; -->(x-<!--\; -\; -->3)$

Algebra of Exponential and Log Functions
This question may have a simple answer or a very complex one, but I am interested in what the reasons are for logarithms and exponential functions having the properties they have. To my knowledge logarithms are the only function that take multiplication into addition and vice versa for the exponential function. I'm sure there must be functions I'm missing.
My question is: What are the algebraic reasons that these functions have these properties and what role does that play in their ubiquity? In addition, resources for further reading (ideally, to the extent possible, without the needing significant background in algebra) would be much appreciated.

Help with differentiation of natural logarithm
Find ${\displaystyle \frac{dy}{dx}}$ given $y=\frac{\ln <!--\; \; -->(8x)}{8x}$
The answer is ${\displaystyle \frac{1-<!--\; -\; -->\ln <!--\; \; -->(8x)}{8x^2}}$
Can you show the process of how this is worked?
Thanks.

Taking the logarithmic derivative of an exponential difference function after applying L'Hospital's Rule
Can somebody please explain the following application of L’Hospital’s Rule?
Find the limit:
$\underset{x\to <!--\; \to \; -->0}{lim}\frac{5^x-<!--\; -\; -->3^x}{x}$
Solution:
Determining that this function has indeterminate form $0/0$, we apply L’Hospital’s rule.
Applying L.H. rule, we get $\underset{x\to <!--\; \to \; -->0}{lim}\frac{5^x\ln <!--\; \; -->5-<!--\; -\; -->3^x\ln <!--\; \; -->3}{1}=\underset{x\to <!--\; \to \; -->0}{lim}(\ln <!--\; \; -->5-<!--\; -\; -->\ln <!--\; \; -->3)=\ln <!--\; \; -->\frac{5}{3}$
The part that I am confused on is the application of the natural logarithm that occurs after the first application of L’Hospital’s Rule, namely $\underset{x\to <!--\; \to \; -->0}{lim}\frac{5^x\ln <!--\; \; -->5-<!--\; -\; -->3^x\ln <!--\; \; -->3}{1}=\underset{x\to <!--\; \to \; -->0}{lim}(\ln <!--\; \; -->5-<!--\; -\; -->\ln <!--\; \; -->3)$. I am not sure how the derivative becomes what it does, nor do I understand the next step. I do understand $\underset{x\to <!--\; \to \; -->0}{lim}(\ln <!--\; \; -->5-<!--\; -\; -->\ln <!--\; \; -->3)=\ln <!--\; \; -->\frac{5}{3}$, however.
Thanks!

If $a²+b²=7ab$ where a and b are positive then show that $log(1/3(a+b))=1/2(loga+logb)$

How to derive this simple equality?
Let us define
$L_i\triangleq <!--\; \triangleq \; -->\log <!--\; \; -->\left({\displaystyle \frac{Prob(x_i=+1)}{Prob(x_i=-<!--\; -\; -->1)}}\right)$
$E\{x_i\}\triangleq <!--\; \triangleq \; -->Prob(x_i=+1)-<!--\; -\; -->Prob(x_i=-<!--\; -\; -->1)$
I need to show that
$E\{x_i\}=\tanh <!--\; \; -->(L_i/2)$
I can write
$\begin{array}{}\text{(1a)}& E\{x_i\}& =e^{log(Prob(x_i=+1)-<!--\; -\; -->Prob(x_i=-<!--\; -\; -->1))}\text{(1b)}& & =e^{L_i}\end{array}$
if I use the following:
$\tanh <!--\; \; -->(z)={\displaystyle \frac{\sinh <!--\; \; -->(z)}{\cosh <!--\; \; -->(z)}}={\displaystyle \frac{e^z-<!--\; -\; -->e^{-<!--\; -\; -->z}}{e^z+e^{-<!--\; -\; -->z}}}={\displaystyle \frac{e^{2z}-<!--\; -\; -->1}{e^{2z}+1}}$
and put $z=L_i/2$, then
$\begin{array}{}\text{(2)}& \tanh <!--\; \; -->(L_i/2)={\displaystyle \frac{e^{L_i}-<!--\; -\; -->1}{e^{l_i}+1}}\end{array}$
how do I arrive at (2) using (1a)

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document How do you solve this equation: $10=2^x+x$
Is it possible to solve this equation?

If a and b are known, how do you find x?

How is this limit being solved? I can't grasp it
I am going over limits for my finals as I notice this example in my schoolbook discribing limits of the undefined form $\frac{0}{0}$ in the shape of an irrational fraction.
$\underset{x\to <!--\; \to \; -->1}{lim}\frac{\sqrt[3]{x}-<!--\; -\; -->1}{\sqrt{2x-<!--\; -\; -->1}-<!--\; -\; -->1}=\frac{0}{0}=\underset{x\to <!--\; \to \; -->1}{lim}\frac{(\sqrt[3]{x}-<!--\; -\; -->1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)(\sqrt{2x-<!--\; -\; -->1}+1)}{(\sqrt{2x-<!--\; -\; -->1}-<!--\; -\; -->1)(\sqrt{2x-<!--\; -\; -->1}+1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}$
I really don't get what is going on here. I know that you are supposed to multiply both the nominator and the denominator by the added value of either. But it appears that here it has been mutliplied by both? And where does this term $(\sqrt[3]{x^2}+\sqrt[3]{x}+1)$ come from?
There is no worked out example, except for the solution after this which is $\frac{1}{3}$ but thats it.
Could anyone please explain this to me? I would be really gratefull! Also sorry if I didn't use the correct terms here and there, not a native english speaker.
Thanks in advance.

Proof $e^x=\exp <!--\; \; -->(x)$?
Define
$\ln <!--\; \; -->(x)={\int <!--\; \int \; -->}_1^x\frac{1}{t}$
Assume I have proven that $\ln <!--\; \; -->x$ is one-to-one and therefore has an inverse $\exp <!--\; \; -->(x)$
Define $e$ as:
$\ln <!--\; \; -->e=1$
Now, if you have no other notion of exponentials, or logarithms, how could define what $e^x$ means and show that its the inverse of $\ln <!--\; \; -->x$?
You are allowed to assume the logarithmic product and quotient property.
Thanks for the help.

How do you solve this exponential equation?
$3(16{)}^x+2(81{)}^x=5(36{)}^x$
How do you change the bases to combine the terms? The correct answer should be 0 and 0.5.
Edit: So this equation can't be solved algebraically? I have to use creative logic to solve it?

What is natural logarithm of this expression?
$y=4\ast <!--\; \ast \; -->[x^9\ast <!--\; \ast \; -->x^6]$
Is it
$\ln <!--\; \; -->(y)=4\ast <!--\; \ast \; -->[9\ln <!--\; \; -->(x)+6\ln <!--\; \; -->(x)]$
or
$\ln <!--\; \; -->(y)=\ln <!--\; \; -->(4)+9\ln <!--\; \; -->(x)+6\ln <!--\; \; -->(x)$

Showing $\frac{x}{1+x}<\log <!--\; \; -->(1+x)<x$ for all $x>0$ using the mean value theorem
I want to show that
$\frac{x}{1+x}<\log <!--\; \; -->(1+x)<x$
for all $x>0$ using the mean value theorem. I tried to prove the two inequalities separately.
$\frac{x}{1+x}<\log <!--\; \; -->(1+x)\iff <!--\; \iff \; -->\frac{x}{1+x}-<!--\; -\; -->\log <!--\; \; -->(1+x)<0$
Let
$f(x)=\frac{x}{1+x}-<!--\; -\; -->\log <!--\; \; -->(1+x).$
Since
$f(0)=0$
and
$f^{\prime }(x)=\frac{1}{(1+x{)}^2}-<!--\; -\; -->\frac{1}{1+x}<0$
for all $x>0$, $f(x)<0$ for all $x>0$. Is this correct so far?
I go on with the second part: Let
$f(x)=\log <!--\; \; -->(x+1)$. Choose $a=0$ and $x>0$ so that there is, according to the mean value theorem, an $x_0$ between $a$ and $x$ with
$f^{\prime }(x_0)=\frac{f(x)-<!--\; -\; -->f(a)}{x-<!--\; -\; -->a}\iff <!--\; \iff \; -->\frac{1}{x_0+1}=\frac{\log <!--\; \; -->(x+1)}{x}$
Since
$x_0>0\Rightarrow <!--\; \Rightarrow \; -->\frac{1}{x_0+1}<1.$
$\Rightarrow <!--\; \Rightarrow \; -->1>\frac{1}{x_0+1}=\frac{\log <!--\; \; -->(x+1)}{x}\Rightarrow <!--\; \Rightarrow \; -->x>\log <!--\; \; -->(x+1)$