Solve for x using log50=600e^{-0.4x}

Find smallest positive integer solution of $x$ for $a^x\le b$
$a\in \mathbb{R},0<a<1.$
$b\in \mathbb{R},0<b<1.$
$x\in \mathbb{N}.$
Let $a^x\le b$
How can I find smallest positive integer solution of $x$ for given $a$ and $b$?

Find the ${\log }_y(x)$given x and y in the form of a number.

So I was doing logs trying to reteach myself pre-calculus and I noticed I don't remember how to turn logs in to numbers. I tried to calculate the ${\log }_{4}8$ so I turned it in to $4^x=8$ and then realized that $8=4^1\sqrt{4}=4^{1.5}$. Isn't there a method to calculate the answer even when it is an irrational number?

Use the properties oflogarithms to simplify the expression.
${\displaystyle 3^{x+1}=27}$

Product of logarithms, prove this identity.

$2\log (a)\log (b)=\log (ab{)}^2-\log (a{)}^2-\log (b)2$​​​​​​​

for a>1and b>1?

What is solution of this equation
I am new to logarithms, I am confused what to do when any constant is raised to a power such as ${\displaystyle 1+\log x}$ .
As in this equation, ${\displaystyle 9^{1+\log x}-3^{1+\log x}-210=0}$
While I try to take log on both sides, things get messy.
All the logs have base 3.

logarithmic derivative of $x^{e^{(x^2+\cos x)}}$
I'm having a hard time taking the derivative of
$f(x)=x^{e^{(x^2+cosx)}}.$
I'm aware that I have to take the logarithm of both sides.
$\ln (y)=\ln (x^{e^{(x^2+\cos x)}})=\ln (x)\cdot e^{(x^2+\cos x)}$
Which I tried to untie, so lets start: First I use the product rule:
$\frac{1}{y}\cdot y^{\prime }=\frac{1}{x}\cdot e^{(x^2+\cos x)}+\ln (x)\cdot e^{(x^2+\cos x)}$
Next the power rule:
$\frac{1}{y}\cdot y^{\prime }=\frac{1}{x}\cdot e^{(x^2+\cos x)}+\ln (x)\ast e^{(x^2+cosx)}\cdot x^2+\cos x\cdot e^{(x^2+\cos x)-1}$
Then I bring the y to the right:
$y^{\prime }=y\cdot (\frac{1}{x}\cdot e^{(x^2+\cos x)}+\ln (x)\cdot e^{(x^2+\cos x)}\cdot x^2+\cos x\cdot e^{(x^2+\cos x)-1}\cdot 2x-\sin x)$
And exchange y with the term:
$y^{\prime }=x^{e^{(x^2+\cos x)}}\cdot (\frac{1}{x}\cdot e^{(x^2+\cos x)}+\ln (x)\cdot e^{(x^2+\cos x)}\cdot x^2+\cos x\cdot e^{(x^2+\cos x)-1}\cdot 2x-\sin x)$
This is extremely overwhelming for me and I have absolutely no clue if this is right, I looked in to the result of wolfram and It doesn't seem to be correct. Any help would be appreciated.

Simplify expression using the properties of logarithms.
${\displaystyle 3{\log }_4{r}^2-2{\log }_{4}r}$