Expert Answers to Algebra 2 Problems

How many numbers less than $x$ have a prime factor that is not $2$ or $3$
I am trying to figure out the number of integers greater than 1 and less than or equal to x that have a prime factor other than 2 or 3. For example, there are only two such integer less than or equal to 7.
It is straight forward to determine how many many integers less than or equal to x have a prime factor other than 2:
$x-<!--\; -\; -->\lfloor {\log }_{2}x\rfloor$
Or to make the same determination about 3:
$x-<!--\; -\; -->\lfloor {\log }_{3}x\rfloor$
What is the method or formula for figuring out how many integers less than or equal to x have a prime factor other than 2 or 3?
I know that it is less than:
$x-<!--\; -\; -->\lfloor {\log }_{2}x\rfloor -<!--\; -\; -->\lfloor {\log }_{3}x\rfloor$
and greater than:
$x-<!--\; -\; -->\lfloor {\log }_{2}x\rfloor -<!--\; -\; -->\lfloor {\log }_{3}x\rfloor -<!--\; -\; -->\lfloor \frac{x}{6}\rfloor$
Thanks

Using logarithms to solve the following equation to find x
$9^{2x}={27}^{1-<!--\; -\; -->x}$ ?? I'm really struggling with this questions. I appreciate your help and if you can please show me your working out so I can understand it too

Solve logarithmic equation for $x$ to find the inverse of $f(x)=\ln <!--\; \; -->(x+\sqrt{x^2+1})$

Find the solution to the differential equation
Assume $x><!--\; >\; -->0$ and let
$x(x+1)\frac{du}{dx}=u^2,$
$u(1)=4.$
I started off by doing some algebra to get:
$\frac{1}{u^2}du=\frac{1}{x^2+x}dx.$
I then took the partial fraction of the right side of the equation:
$\frac{1}{u^2}du=(\frac{1}{x}-<!--\; -\; -->\frac{1}{x+1}).$
I then took the integral of both sides:
$-<!--\; -\; -->\frac{1}{u}=\log <!--\; \; -->x-<!--\; -\; -->\log <!--\; \; -->(x+1)+C.$
From here I don't know what to do because we are solving for $u(x)$ and I'm not sure how to get that from $-<!--\; -\; -->\frac{1}{u}$

How to find explicit formula for a sequence seems to be geometric but isn't?
I recently came across this interesting problem, where the progression was neither arithmetic nor geometric. The problem asks for an explicit formula for the following recursive formula:
$a_1=2$
$a_n=5(a_{n-<!--\; -\; -->1}+2)$
So the first five terms are 2,20,110,560,2810.
I tried to distribute the 5 in the second equation to get $5a_{n-<!--\; -\; -->1}+10$ and then tried to apply the formula for a geometric sequence, which is $a_1(r{)}^{n-<!--\; -\; -->1}$, but got only $2(5{)}^{n-<!--\; -\; -->1}$, which clearly doesn't work for the sequence.
I also tried using finite differences on the first five terms to see if it was just a polynomial rule, but that didn't work.
How would you go about doing a problem like this?

Dealing with Logarithms. $\log <!--\; \; -->(b^x+a)=\log <!--\; \; -->(c)$
What methods/techniques are available to solve for x in the following type of situation:
$\log <!--\; \; -->(b^x+a)=\log <!--\; \; -->(c)$
The only log methods I have been exposed to are using the power laws and bring x out, which you cannot do in this case.
Thanks for your help.

Modeling the relationship between perimeter and area
Is there any equation that models the relationship between the area and perimeter of a rectangle?

How do I evaluate this without using taylor expansion :$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{x}^2\log <!--\; \; -->(\frac{x+1}{x})-<!--\; -\; -->x$?
How do I evaluate this without using Taylor expansion?
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{x}^2\log <!--\; \; -->\left(\frac{x+1}{x}\right)-<!--\; -\; -->x$
Note: I used Taylor expansion at $z=0$ and I have got $\frac{-<!--\; -\; -->1}{2}$
Thank you for any help

What are the asymptotes of y=3/(x−1)−2 and how do you graph the function?

Solve.
11.6 is 58% of what number?

Solving logarithmic equation, different bases
What number do I need to multiply both sides with? I have worked for an hour on this but it is the first time I am using this website so it is impossible for me to write what I have already done. If you can give me tips on how to solve it I would appreciate it a lot!
$2{\log }_6<!--\; \; -->(\sqrt{x}+\sqrt[4]{x})={\log }_4<!--\; \; -->x$

Modeling a greatest integer function
I'm trying to model a function that resembles a greatest integer function. The domain is from [0, $\mathrm{\infty <!--\; \infty \; -->}$). The inputs from 0 to 1.5 (non-inclusive) need to be mapped to an output of 0, and 1.5 to $\mathrm{\infty <!--\; \infty \; -->}$ mapped to 1. But, I'm trying to not use a piecewise function. Is it possible to accomplish this?
Here's what I've tried:
$f(x)=\lfloor \frac{x}{1.5}\rfloor$

Prove that there exists an integer greater than x such that any polynomial f(x) will be strictly non-negative and get large?
I am taking a number theory class and so far I have been proving modular congruences, modular arithmetic, and prime properties. There is this theorem that came up in the textbook and apparently it does not involve any modular arithmetic. The theorem is as follows
Suppose $f(x)=a_n{x}^n+a_{n-<!--\; -\; -->1}{x}^{n-<!--\; -\; -->1}+\cdots <!--\; \cdots \; -->+a_0$ is a polynomial of degree $n>0$. Then there exists an integer k such that if $x>k$ then $f(x)>0$.
I feel like this would come up in real analysis, but I have not come that far in my studies. I have an idea of applying induction and using the ceiling function somehow, but I have no idea how to start off this proof. Any help will do and thank you

Log Log Integrals II
The integral
$\begin{array}{r}{I}_4={\int <!--\; \int \; -->}_0^1\ln <!--\; \; -->(1-<!--\; -\; -->x){\ln }^2<!--\; \; -->(\ln <!--\; \; -->\left(\frac{1}{x}\right))\frac{dx}{x}\end{array}$
can be expressed as
$\begin{array}{r}{I}_4={\mathrm{\zeta <!--\; \zeta \; -->}}^{{}^{″}}(2)-<!--\; -\; -->\frac{{\mathrm{\gamma <!--\; \gamma \; -->}}^2{\mathrm{\pi <!--\; \pi \; -->}}^2}{6}-<!--\; -\; -->\frac{\mathrm{\gamma <!--\; \gamma \; -->}{\mathrm{\pi <!--\; \pi \; -->}}^2}{3}\ln <!--\; \; -->\left(\frac{2\mathrm{\pi <!--\; \pi \; -->}}{A^{12}}\right)+\frac{{\mathrm{\pi <!--\; \pi \; -->}}^4}{36}\end{array}$
where $A$ is the Glaisher-Kinkelin constant.
The integrals
$\begin{array}{r}{I}_5={\int <!--\; \int \; -->}_0^1\frac{\ln <!--\; \; -->(1-<!--\; -\; -->x)}{\sqrt{\ln <!--\; \; -->\left(\frac{1}{x}\right)}}{\ln }^2<!--\; \; -->(\ln <!--\; \; -->\left(\frac{1}{x}\right))\frac{dx}{x}\end{array}$
and
$\begin{array}{r}{I}_6={\int <!--\; \int \; -->}_0^1\frac{\ln <!--\; \; -->(1-<!--\; -\; -->x)}{{\ln }^{3/2}<!--\; \; -->\left(\frac{1}{x}\right)}{\ln }^2<!--\; \; -->(\ln <!--\; \; -->\left(\frac{1}{x}\right))\frac{dx}{x}\end{array}$
can be expressed in terms of ${\mathrm{\partial <!--\; \partial \; -->}}_s^2\mathrm{\zeta <!--\; \zeta \; -->}(s){|}_{s=3/2}$ and ${\mathrm{\partial <!--\; \partial \; -->}}_s^2\mathrm{\zeta <!--\; \zeta \; -->}(s){|}_{s=1/2}$, respectively. Can these integrals be evaluated in closed form expression without the use of derivatives of the Zeta function?
If the integrals can be evaluated in such a way what is the resulting value?

How to solve this exponential equation? $2^{2x}{3}^x=4^{3x+1}$
I haven't been able to find the correct answer to this exponential equation:
$\begin{array}{rl}{2}^{2x}{3}^x& =4^{3x+1}\\ 2^{2x}{3}^x& =2^2\times <!--\; \times \; -->2^x\times <!--\; \times \; -->3^x\\ 4^{3x+1}& =4^3\times <!--\; \times \; -->4^x\times <!--\; \times \; -->4\\ 6^x\times <!--\; \times \; -->4& =4^x\times <!--\; \times \; -->256\\ x{\log }_6<!--\; \; -->6+{\log }_6<!--\; \; -->4& =x{\log }_6<!--\; \; -->4+{\log }_6<!--\; \; -->256\\ x+{\log }_6<!--\; \; -->4& =x{\log }_6<!--\; \; -->4+{\log }_6<!--\; \; -->256\\ x-<!--\; -\; -->x{\log }_6<!--\; \; -->4& ={\log }_6<!--\; \; -->256-<!--\; -\; -->{\log }_6<!--\; \; -->4\\ x(1-<!--\; -\; -->{\log }_6<!--\; \; -->4)& ={\log }_6<!--\; \; -->256-<!--\; -\; -->{\log }_6<!--\; \; -->4\\ x& ={\displaystyle \frac{{\log }_6<!--\; \; -->256-<!--\; -\; -->{\log }_6<!--\; \; -->4}{1-<!--\; -\; -->{\log }_6<!--\; \; -->4}}\\ x& =10.257\end{array}$
so when I checked the answer I wasn't able to make them equal, I have tried variants of this method but I feel I'm missing something..

I need help modeling the velocity profile of a river's current
I am trying to solve Zermelo's Navigation Problem.
One of the cases I'm looking at is when the river's current is a function of the x-position only.
From what I learned in Fluid Mechanics courses, I know that at the two ends when (i.e. the river banks) the velocity should be zero. Then in the center the velocity is at its maximum value.
In other words: $v(x=0)=v(x=L)=0$, and $v(x=0.5L)=V_{\text{max}}$
Everything I learned in the past was these velocities as function of radius, which makes sense for pipes and tubes, but since this can be thought of a 2D rectangular flow, I can't figure this out.
I know it should be a quadratic expression.

A general theory of convolution product
in my childhood, I learned about convolution products for function over R (1). For quite a while now, I have played with polynomial rings, where also, the product is sometime called a convolution product (2). I am now discovering convolution products for arithmetic function and multiplicative functions [Apostol, Introduction to Analytic Number Theory] (3).
In case (1), the domain is R, an ordered abelian group. In case (2), seeing a ring of polynomial as a group ring Z⟨G⟩, the domain is an abelain group, without ordering. In case (3), the domain seems to be the monoid $(\mathbb{Z},\times <!--\; \times \; -->)$, with an ordered by the divisibilty relation.
I don't have anything very formal, but it seems to me that there should be a general theory of convolution I don't know yet about ? Is the monoid structure the most general domain, or maybe something less structured as an acyclic graph ? Would you have lectures notes on such a theory ?

If $e^{2\mathrm{\pi <!--\; \pi \; -->}i}=1$ then is $2\mathrm{\pi <!--\; \pi \; -->}i=0$?
So with Euler identity I used $2\mathrm{\pi <!--\; \pi \; -->}(\mathrm{\tau <!--\; \tau \; -->})$ instead and got $e^{2\mathrm{\pi <!--\; \pi \; -->}i}=1$, and took natural logarithm $2\mathrm{\pi <!--\; \pi \; -->}i=0$? But is see online the answer is $\mathrm{6.28....}i$.

Solution of an exponential equation
Probably very simple question. Why the solution of
$1=n(1-<!--\; -\; -->a{)}^t$
in terms of t is equal to:
$t=\frac{\ln <!--\; \; -->n}{\ln <!--\; \; -->\frac{1}{1-<!--\; -\; -->a}}$

Use logarithms to solve the equation for t.
$9-<!--\; -\; -->e^{0.2t}=5$