Expert Answers to Algebra 2 Problems

Proof of logarithmic identity ${\log }_g<!--\; \; -->x={\log }_a<!--\; \; -->x\cdot <!--\; \cdot \; -->{\log }_g<!--\; \; -->a$
I have to prove the alleged link between the logarithms in base g and a
${\log }_g<!--\; \; -->x={\log }_a<!--\; \; -->x\cdot <!--\; \cdot \; -->{\log }_g<!--\; \; -->a$
I know that this can be written as:
$\frac{\ln <!--\; \; -->x}{\ln <!--\; \; -->g}=\frac{\ln <!--\; \; -->x}{\ln <!--\; \; -->g}$
But does this prove the alleged link or do i have to do something further?
I know that this is derived from:
${\log }_g<!--\; \; -->x=\frac{\ln <!--\; \; -->x}{\ln <!--\; \; -->g}$

How to solve this kind of equation $(x^y=y^x)$$(x^y=y^x)$
I'm little bit stuck with this system of equations :
$x^y=y^x$ and $x^3=y^2$
An obvious solution is $(x,y)=(1,1)$ but what about the solution $(9/4,27/8)$?
I know the relation $a^r=e^{r\log <!--\; \; -->a}$ but it doesn't help me.
Thanks

The polynomial
$y=-<!--\; -\; -->0.79x^4+3x^3+27.3$
describes the billions of flu virus perticles in a person's boby x days after being infected. Find the number of virus particles, in billions, after 1 day

How does exponentiation relate to multiplication?
My book derives the logarithm function as a definite integral of $1/x$ and defines the exponential function as its inverse. It then extends this definition to other bases:
$b^x=e^{\ln <!--\; \; -->(b)x}$
But then somehow we get this really nice property:
$b^x=\underset{x\text{ times}}{\underset{⏟<!--\; ⏟\; -->}{b\cdot <!--\; \cdot \; -->b\cdot <!--\; \cdot \; -->\dots <!--\; \dots \; -->\cdot <!--\; \cdot \; -->b}}$
if $x$ is an integer.
This property is too nice to just be a coincidence. Is there a deeper reason this property emerges? I mean it's hard to see how functions derived from relatively complex operations on $1/x$ (integration) would somehow simplify to such a simple definition for integer arguments.

What log rule was used to simply this expression?
I'm unclear how the left side is equal to the right side.
$365\log <!--\; \; -->(365)-<!--\; -\; -->365-<!--\; -\; -->305\log <!--\; \; -->(305)+305-<!--\; -\; -->60\log <!--\; \; -->(365)=305\log <!--\; \; -->\left(\frac{365}{305}\right)-<!--\; -\; -->60$
I know $\log <!--\; \; -->(a)-<!--\; -\; -->\log <!--\; \; -->(b)=\log <!--\; \; -->(a/b)$ but if you stick constants before each ln() then how do you apply the rule to get 305 as the constant on the right side of the equation?

Determining $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{(n^{\frac{1}{n}}-<!--\; -\; -->1)}^n$ with only elementary math
I am trying to find this limit:
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{(n^{\frac{1}{n}}-<!--\; -\; -->1)}^n,$
I tried using exponential function, but I see no way at the moment. I am not allowed to use any kind of differentiation or other topics of advanced math, only induction and school math are possible. Thank you

Logarithmic function with strange bases
Given ${\log }_{4n}<!--\; \; -->40\sqrt{3}={\log }_{3n}<!--\; \; -->45$, find n.
I have rewritten ${\log }_{3n}<!--\; \; -->45$ as ${\displaystyle \frac{{\log }_{4n}<!--\; \; -->45}{{\log }_{4n}<!--\; \; -->3n}}$ and multiplied to get
${\log }_{4n}<!--\; \; -->40\sqrt{3}\cdot <!--\; \cdot \; -->{\log }_{4n}<!--\; \; -->3n={\log }_{4n}<!--\; \; -->45$
but do not know how to continue. Hints would be greatly appreciated, but please don't give me the answer. Thank you.

Calculate ${\int <!--\; \int \; -->}_0^1\frac{{\log }^2<!--\; \; -->x\log <!--\; \; -->(1+x^2)}{1-<!--\; -\; -->x^2}dx$
I found $-<!--\; -\; -->{\displaystyle \frac{{\mathrm{\pi <!--\; \pi \; -->}}^4}{32}}+2G^2+{\displaystyle \frac{7}{4}}\zeta (3)\log <!--\; \; -->2$ where $G$ is the Catalan's constant.

True or false logarithmic branches
Say whether the following are true or false. Give a short proof.
1) $log(-<!--\; -\; -->z)+i\mathrm{\pi <!--\; \pi \; -->}$ is a branch of the logarithmic function whose branch cut is the non-negative real axis
2)If $g(z)$ is a branch of the logarithmic function with domain D and $h(z)$ is a branch analytic in D, then there is an integer $m$ with $h(z)=g(z)+2m\mathrm{\pi <!--\; \pi \; -->}i$
Thanks

3.24 divided by 82 estimate the quotion

At sea level, air pressure is 30 inches of mercury. At an altitude of h feet above the sea level, the air Pressure, P, in inches of mercury, is given by the function
$P=30e^{-<!--\; -\; -->0.0000323h}$
Determine the rate of change of the air pressure, in inches of mercury per feet, for a pilot of a small plane passing through an altitude of 1300 feet. Round your answer to five decimal places.

Melissa and Emily are playing at the pool.They have three different measuring jars for; liters, cups, and pints. They find that 7 cups of water and 3 liters of water filled 9.8 pints. Later, they find that if they start with 5 liters of water and remove 9 cups of water, they end up with 6 pints of water. Model the given two situations as a system of linear equations. Based on your equations, determine the relationship between;
●cups and liters,
●cups and pints,
●and pints and liters.
Show all your work and submit your work along with the resulting relationships.

Proof ' that $\ln <!--\; \; -->(x)$ converges
Where is the flaw in the following 'proof '?
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\left[\frac{\mathrm{d}}{\mathrm{d}x}\{\ln <!--\; \; -->(x)\}\right]=\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\left[\frac{1}{x}\right]=0⟹<!--\; ⟹\; -->\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}[\ln <!--\; \; -->(x)]=\text{constant}\in <!--\; \in \; -->\mathbb{R},$
therefore $\ln <!--\; \; -->(x)$ converges to some real number. $\mathrm{◻<!--\; ◻\; -->}$

Trouble with Logarithmic Differentiation
Hey guys I'm trying to find the derivative of this equation using logarithmic differentiation but I'm having some trouble. Wolfram Alpha is giving me different answers and I'm having difficulty understanding the steps it spits out. Could somebody take a look at what I'm doing and tell me where I went wrong?
The problem I need to differentiate is:
$y=(\sin <!--\; \; -->x{)}^{\ln <!--\; \; -->x}$
My first step is to take the logarithm of both sides and factor out the exponent.
$\ln <!--\; \; -->(y)=\ln <!--\; \; -->x\ast <!--\; \ast \; -->\ln <!--\; \; -->(\sin <!--\; \; -->x)$
Next I try to differentiate both sides, using the rules for logarithms:
$\frac{1}{y}\ast <!--\; \ast \; -->y^{\prime }=\frac{1}{x}\frac{1}{\sin <!--\; \; -->x}\cos <!--\; \; -->x\Rightarrow <!--\; \Rightarrow \; -->\frac{1}{y}\ast <!--\; \ast \; -->y^{\prime }=\frac{\cos <!--\; \; -->x}{x\sin <!--\; \; -->x}$
Which finally renders:
$\frac{dy}{dx}=\frac{\cos <!--\; \; -->x}{x\sin <!--\; \; -->x}(\sin <!--\; \; -->x{)}^{\ln <!--\; \; -->x}$
But this doesn't seem to be the right answer, could somebody help me out?

y=x-4
y=-2x+5
find x

How to solve exponential inequality with $x$
I need to solve the following inequality.
$\ln <!--\; \; -->(x)-<!--\; -\; -->x>0.$
I oddly remember that it can only be done by using the graph... Is it true?
I have the same problem with
$e^x(x-<!--\; -\; -->1)>-<!--\; -\; -->2.$
Thanks!

Number raised to log expression
I am struggling with what I think should be some a basic log problem:
Show that $3^{lo{g}_{2}n}=n^{lo{g}_{2}3}$
I know that $3^{lo{g}_{3}n}=n$ and $lo{g}_{2}n=lo{g}_{3}n/lo{g}_{3}2$
I was attempting something similar to:
$3^{lo{g}_{3}n/lo{g}_{3}2}=3^{lo{g}_{3}n-<!--\; -\; -->lo{g}_{3}2}$ but then I got stuck. Am I on the right track by using the change of base and then subtracting?
EDIT: I'm not trying to exactly 'show that' the two expressions are equal. I am looking to figure out how to simplify the expression on the left to the one on the right

Determine the form of a particular solution to $L(y)=\mathrm{\varphi <!--\; \varphi \; -->}(x)$ for $\mathrm{\varphi <!--\; \varphi \; -->}(x)$ as given if the solution to the associated homogeneous equation L(y)=0 is $y_h=c_1{e}^{2x}+c_2{e}^{3x}$
$$\begin{gathered} 1)\mathrm{\varphi <!--\; \varphi \; -->}(x)=2x-<!--\; -\; -->7 \\ 2)\mathrm{\varphi <!--\; \varphi \; -->}(x)=-<!--\; -\; -->3x^2 \\ 3)\mathrm{\varphi <!--\; \varphi \; -->}(x)=4e^{2x} \\ 4)\mathrm{\varphi <!--\; \varphi \; -->}(x)=2\cos <!--\; \; -->(3x) \end{gathered}$$

Determine which of the numbers are solutions to the given equation.
$5(x-<!--\; -\; -->1)=4x+2;x=2,x=7,x=-<!--\; -\; -->1$
Is x=2 a solution to the equation $5(x-<!--\; -\; -->1)=4x+2?$

Solve ${\log }_2<!--\; \; -->(1+\frac{1}{x-<!--\; -\; -->1})<1$
I don't get how my teacher got two different equations out of the one. One is $>0$ and the other one is $<2$. Be detailed please.