Learn How to Graph Logarithmic Equations

Characteristic function of logarithm of random variable
If I know the characteristic function ${\mathrm{\varphi <!--\; \varphi \; -->}}_X(t)$ of a random variable $X>0$, how can I write the characteristic function ${\mathrm{\varphi <!--\; \varphi \; -->}}_Y(t)$ of $Y=\log <!--\; \; -->(X)$?
I know that ${\mathrm{\varphi <!--\; \varphi \; -->}}_X(t)=E[e^{itX}]$ and ${\mathrm{\varphi <!--\; \varphi \; -->}}_Y(t)=E[e^{it\log <!--\; \; -->(X)}]$. But I can't derive one from the other. Any idea? I would like to use ${\mathrm{\varphi <!--\; \varphi \; -->}}_X(t)$ to calculate the second moments $X$

Solve $\ln <!--\; \; -->(x)+\ln <!--\; \; -->(x-<!--\; -\; -->1)=0$ for x
Solve the following equation for x;
$\ln <!--\; \; -->(x)+\ln <!--\; \; -->(x-<!--\; -\; -->1)=0$
What I did is the following but I'm pretty sure its wrong..
$\ln <!--\; \; -->(x)+\ln <!--\; \; -->(x-<!--\; -\; -->1)=0$
$\ln <!--\; \; -->(x)=-<!--\; -\; -->\ln <!--\; \; -->(x-<!--\; -\; -->1)$
$e^{\ln <!--\; \; -->(x)}=e^{-<!--\; -\; -->\ln <!--\; \; -->(x-<!--\; -\; -->1))}$
$x=-<!--\; -\; -->x-<!--\; -\; -->1$
$2x=-<!--\; -\; -->1$
$x=-<!--\; -\; -->\frac{1}{2}$

Exponential and Logarithmic Differentiation.
Q. If $x{e}^{xy}=y+{\sin }^2<!--\; \; -->x$, then find $\frac{dy}{dx}$ at x=0.
If we differentiate the function directly as follows:
$e^{xy}+x{e}^{xy}[y+x\frac{dy}{dx}]=\frac{dy}{dx}+\sin <!--\; \; -->\left(2x\right)$
At x=0 We have
$1+0=\frac{dy}{dx}+0$ which gives
$\frac{dy}{dx}=1$
But if we apply logarithm and differentiate we have:
$\log <!--\; \; -->\left(x\right)+xy=\log <!--\; \; -->(y+{\sin }^2<!--\; \; -->x)$ which when differentiated gives
$\frac{1}{x}+y+x\frac{dy}{dx}=\frac{\frac{dy}{dx}+\sin <!--\; \; -->2x}{y+{\sin }^2<!--\; \; -->x}$
If we take x=0 here, we get undefined values. And we are unable to obtain the value of dy/dx individually. Is there any mistake in the logarithmic method?

Solving $A{e}^x=Bx$ analytically, where $A$ and $B$ are constants?
This equation mixes both exponential terms and linear terms, something which I do not know how to deal with. Any pointers?

Define logarithmic decrement

applying logarithm law question
Here is my equation (below) on which I am applying log
$X=\frac{a}{b}(c-<!--\; -\; -->d)$
so far I applied it as
$\log <!--\; \; -->X=\log <!--\; \; -->(a)-<!--\; -\; -->\log <!--\; \; -->(b)+[\log <!--\; \; -->\left(c\right)-<!--\; -\; -->\log <!--\; \; -->\left(d\right)]$

How to define this logarithmic function
I am trying to get my head around the definition of this function (that I concocted as an exercise in defining a function). Let $f$ denote the function satisfying:
$f(0)=+\mathrm{\infty <!--\; \infty \; -->}$, and $f(+\mathrm{\infty <!--\; \infty \; -->})=0$ and is a logarithmic curve.
I know this should be elementary to define but I am stumped.

Finding unknown in log equation
I was given a log equation:
$D=10\log <!--\; \; -->(I/I_0)$
$I$ is the unknown in this case, $I_0={10}^{-<!--\; -\; -->12}$ and $D=89.3$
I did the following steps:
$\begin{array}{rl}89.3& =10\log <!--\; \; -->\left(\frac{I}{{10}^{-<!--\; -\; -->12}}\right)\\ \frac{89.3}{10}& =\log <!--\; \; -->\left(\frac{I}{{10}^{-<!--\; -\; -->12}}\right)\\ 8.93& =\log <!--\; \; -->\left(\frac{I}{{10}^{-<!--\; -\; -->12}}\right)\end{array}$
I'm not quite sure how to isolate I after step 3, and I'm also unsure if dividing $89.3/10$ is correct as well. So how can I find the unknown ($I$)?

Why does the same inequality give different answers?
$({\log }_2<!--\; \; -->\left(x\right)-<!--\; -\; -->2)({\log }_2<!--\; \; -->\left(x\right)+1)<0$
has a solution $\frac{1}{2}<x<4$
But when we take the second part alone that is
$({\log }_2<!--\; \; -->\left(x\right)+1)<0$
it gives a solution $0<x<\frac{1}{2}$ why is $x>\frac{1}{2}$ in the first case but $x<\frac{1}{2}$ in the second case?

Finding time constants of a circuit?
So this is a homework question and I am having trouble figuring out what they are asking.
'The potential difference (voltage) across the capacitor at time t > 0 is given by $V_C(t)=q(t)/C$. The quantity RC has the dimensions of time and is often called the time constant for the circuit. How many time constants does it take for a capacitor to charge to 90% of the applied voltage, V0? Justify your answer'
So change in V, or $V_C$, or $\mathrm{\delta <!--\; \delta \; -->}V$ is 90%. In other words we have $0.9V=q(t)/C$?
I have found in a previous question that $q(t)=V_{0}C(1-<!--\; -\; -->e^{-<!--\; -\; -->\frac{t}{CR}})$ so
$0.9V=\frac{V_{0}C(1-<!--\; -\; -->e^{-<!--\; -\; -->\frac{t}{CR}})}{C}$
However I am not sure where to go from here, if I am even on the right track at all.

Express ${\log }_5<!--\; \; -->288$ in terms of decimal logarithms $\log <!--\; \; -->2$ and $\log <!--\; \; -->3$
Assuming $a=\log <!--\; \; -->2$ and $b=\log <!--\; \; -->3$ (log is the base 10 logarithm). I have to find ${\log }_5<!--\; \; -->288$. How can I do this?
I've tried transforming $\log <!--\; \; -->2$ to $\frac{{\log }_5<!--\; \; -->2}{{\log }_5<!--\; \; -->10}$ and same for $b$. Then it's
$\frac{5{\log }_5<!--\; \; -->2+2{\log }_5<!--\; \; -->3}{{\log }_5<!--\; \; -->10}=\frac{{\log }_5<!--\; \; -->288}{{\log }_5<!--\; \; -->10}=5\log <!--\; \; -->2+2\log <!--\; \; -->5=5a+3b$
Is that correct?

Conformal Mapping Between Two Domains (log)
Does anyone have a recommendation as how to go about solving this problem?
I want a conformal from G to H where
$G=\{z\in <!--\; \in \; -->\mathbb{C}||z|<1,|z+i|>\sqrt{2}\},S=\{z\in <!--\; \in \; -->\mathbb{C}|\mathbb{R}e(z)\in <!--\; \in \; -->(-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->},\mathrm{\pi <!--\; \pi \; -->})\}.$
The previous part of the question was to do with branches of the log. Any advice would be most appreciated!
The sort of answer that I'm looking for is this:
Use f(z) = $\mathrm{\alpha <!--\; \alpha \; -->}i\log <!--\; \; -->(z)+\mathrm{\beta <!--\; \beta \; -->}\arg <!--\; \; -->(z)$
then show how to find $\mathrm{\alpha <!--\; \alpha \; -->}$ and $\mathrm{\beta <!--\; \beta \; -->}$ using the boundaries.
Thanks!

Logarithms with trigonometric inequality
My class is going to have an exam tomorrow, but we can't figure out how to solve such equations.
${\log }_{tg(x)}<!--\; \; -->\sqrt{\sin <!--\; \; -->(x{)}^2-<!--\; -\; -->5/12}<1$
We tried to transform $1$ to ${\log }_{tg(x)}<!--\; \; -->tg(x)$ and solve it as
$\sqrt{\sin <!--\; \; -->(x{)}^2-<!--\; -\; -->5/12}<tg(x)$
But we don't know how to continue. Any help will be greatly appreciated.

help with logarithmic integration.
I've been googling some tutorials on integrating logarithms for my calc 2 class and I've found a lot of good stuff. Unfortunately nothing has answered how to handle a problem that I have. I've tried u-substitution and have still been getting it wrong or just lost. My problem is the $\int <!--\; \int \; -->(\frac{9}{x^3}+\frac{2}{9x})dx$ (sorry, I'm a little new to the math section and don't know how to format that nicely). So far I've tried finding a common denominator which changes the equation to $\frac{81+2x^2}{9x^3}$. I tried doing $u=9x^3$ and $du=27x^{2}dx$ or $\frac{1}{27}du=x^{2}dx$. Now the problem is kind of ugly and I'm not sure what to do with that next. We haven't covered integration by parts yet in this class and I've found that comes up, so I'm not sure if it's possible to evaluate without that technique. I really appreciate any help!
Thank you in advance

calculate ${\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}}{\int <!--\; \int \; -->}_0^x\log <!--\; \; -->(\sin <!--\; \; -->(x-<!--\; -\; -->y))dydx$
I was asked to find the integral ${\iint <!--\; \iint \; -->}_A\log <!--\; \; -->(\sin <!--\; \; -->(x-<!--\; -\; -->y))dxdy$ where $A$ is the triangle $y=0,x=\mathrm{\pi <!--\; \pi \; -->},y=x$ in the first quadrant.
I was given a hint: evaluate ${\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}}\log <!--\; \; -->(\sin <!--\; \; -->(t))dt$ using symmetry.
What I did:
I inferred from the hint that the variable change $t=x-<!--\; -\; -->y$ is the way to go, so $t=x-<!--\; -\; -->y$, and since we integrate by $y$ first, then $x$ is a "constant" and $dt=-<!--\; -\; -->dy$, and since $y$ transitions from $0$ to $x$, then $t$ transitions from $x$ to $0$, so we can rewrite the integral:
${\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}}{\int <!--\; \int \; -->}_0^x\log <!--\; \; -->(\sin <!--\; \; -->(x-<!--\; -\; -->y))dydx={\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}}{\int <!--\; \int \; -->}_x^0-<!--\; -\; -->\log <!--\; \; -->(\sin <!--\; \; -->(t))dtdx={\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}}{\int <!--\; \int \; -->}_0^x\log <!--\; \; -->(\sin <!--\; \; -->(t))dtdx$
And here I am stuck. Firstly, I don't know how ${\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}}\log <!--\; \; -->(\sin <!--\; \; -->(t))dt$ is related to the question, since in the question the limits are $0$ and $x$. not $0$ and $\mathrm{\pi <!--\; \pi \; -->}$. They are not the same thing (even though $x$ transitions from $0$ to $\mathrm{\pi <!--\; \pi \; -->}$).
But even if I did, how would I evaluate ${\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}}\log <!--\; \; -->(\sin <!--\; \; -->(t))dt$??

Summation with Ceilinged Logarithmic Function
According to Johann Blieberger's paper - "Discrete Loops and Worst Case Performance" (1994):
$\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}\lceil {\log }_2<!--\; \; -->(i)\rceil =n\lceil {\log }_2<!--\; \; -->(n)\rceil -<!--\; -\; -->2^{\lceil {\log }_2<!--\; \; -->(n)\rceil }+1$
Now, I was wondering if someone knows what the following may equal?
$\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}i\lceil {\log }_2<!--\; \; -->(i)\rceil =?$

Condensing Fractional Logarithms
Does the following condense to the following:
${\log }_2<!--\; \; -->z+({\log }_2<!--\; \; -->x)/2+({\log }_2<!--\; \; -->y)/2={\log }_2<!--\; \; -->(z\sqrt{x}\sqrt{y})$ or to ${\log }_2<!--\; \; -->(z\sqrt{xy})$?

If $y=1-<!--\; -\; -->x+\frac{x^2}{2!}-<!--\; -\; -->\frac{x^3}{3!}+\dots <!--\; \dots \; -->$ and $z=-<!--\; -\; -->y-<!--\; -\; -->\frac{y^2}{2}-<!--\; -\; -->\frac{y^3}{3}-<!--\; -\; -->\dots <!--\; \dots \; -->$ then $\ln <!--\; \; -->(\frac{1}{1-<!--\; -\; -->e^x})$
I can see that $y=e^{-<!--\; -\; -->x}$ and $z=\ln <!--\; \; -->(1-<!--\; -\; -->y)$. And, so $z=\ln <!--\; \; -->(1-<!--\; -\; -->e^{-<!--\; -\; -->x})$. But how to go further?

Solving ${\sqrt{2}}^x={\sqrt{3}}^x$
I am studying logarithms and exponents. I am not sure how to go about solving this problem. I seem too keep going in circles using the rules of log and exp.
$(\sqrt{2}{)}^x=(\sqrt{3}{)}^x$

Expressing $\ln <!--\; \; -->\sqrt[3]{54}$ in terms of $\ln <!--\; \; -->2$ and/or $\ln <!--\; \; -->3$
I know that $\sqrt[3]{54}={54}^{1/3}$ but otherwise I don't know how to address these types of problems. How do I solve this, and is there a general way of tackling these problems? Like, if I had $\ln <!--\; \; -->.75$ or $\ln <!--\; \; -->\frac{8}{9}$, for example. These types of problems that I have are all different and I don't understand a general way of looking at them.
thanks.