Since we are looking for an inverse function, we start by swapping the x and y variables in our original equation.

$\displaystyle x = \frac{2}{3}e^{4y+1}$

Now we have to solve for y. To do this we have to work towards isolating y. First we remove the constant multiplier:

$\displaystyle \frac{3}{2}x = \frac{3}{2}*\frac{2}{3}e^{4y+1}$

$\displaystyle \frac{3}{2}x = e^{4y+1}$

Next we eliminate the base on the right side by taking the natural log of both sides.

$\displaystyle \ln\frac{3}{2}x = \ln e^{4y+1}$

$\displaystyle \ln\frac{3}{2}x = {4y+1}$

Subtract 1 and divide by 4:

$\displaystyle \ln\frac{3}{2}x - 1 = {4y+1} - 1$

$\displaystyle \ln\frac{3}{2}x - 1 = 4y$

$\displaystyle \frac{1}{4}\left ( \ln\frac{3}{2}x - 1 \right ) = 4y*\frac{1}{4}$

$\displaystyle \frac{1}{4}\ln {\frac{3}{2}x}-\frac{1}{4} = y$

$\displaystyle y = \frac{1}{4}\ln {\frac{3}{2}x}-\frac{1}{4}$

To condense this, the second term must have the 2 in front moved inside.

$\displaystyle \ln{8x}+\ln(3x^2)^2-\ln2$

$\displaystyle \ln8x+\ln9x^4-\ln2$

When adding two logs, multiply their insides; when subtracting two logs, divide their insides.

$\displaystyle \ln((8x)(9x^4))-\ln2$

$\displaystyle \ln72x^5-\ln2$

$\displaystyle \ln\frac{72x^5}{2}$

$\displaystyle \ln36x^5$

To expand a logarithm, quantities in the inside that are multiplied get added and quantities in the inside that are divided get subtracted.

$\displaystyle \log_{3}27+\log_3(x^4)+\log_3(y^3)$

$\displaystyle \log_33^3+4\log_3(x)+3\log_3(y)$

$\displaystyle 3+4\log_3(x)+3\log_3(y)$

The first step to solving this problem is to realize that

$\displaystyle e^{2x}=(e^x)^2$

Then, the equation falls into the follow form which resembles a quadratic.

$\displaystyle (e^x)^2-5e^x+6=0$

Let $\displaystyle y=e^x$. Then,

$\displaystyle y^2-5y+6=0$

$\displaystyle (y-3)(y-2)=0$

Thus, $\displaystyle y=2$ and $\displaystyle y=3$.

Since $\displaystyle y=e^x$,

$\displaystyle 2=e^x\Rightarrow \ln2=\ln{e^x}\Rightarrow x=\ln2$

$\displaystyle 3=e^x\Rightarrow \ln3=\ln{e^x}\Rightarrow x=\ln3$

Using the quotient rule for logarithms we can condense these two logarithms into a single logarithm. 

$\displaystyle \log\left(\frac{72x}{12}\right)$ 

We then obtain our answer by simple division.

Using the properties of logarithms we first simplify the expression to $\displaystyle \ln(y^3)-\ln(x^2y^4)$.  Then we use the quotient rule for logarithms and cancel some terms to obtain our answer.

Using the properties of logarithms we first rewrite the expression as $\displaystyle \log(x^2)-\log(y^3)+\log(x^2y^3)$. Now we combine the three pieces into the form $\displaystyle \log\left(\frac{x^2x^2y^3}{y^3}\right)$. We then obtain our answer when we combine the terms with$\displaystyle x$ and cancel those with $\displaystyle y$.

$\displaystyle 4^{x-2}=64$

$\displaystyle 4^{x-2}=4^3$

$\displaystyle \log_{4}4^{x-2}=\log_{4}4^3$

$\displaystyle x-2=3\Rightarrow x=5$

When multiplying exponents with a common base, you add both the exponents together. Hence, $\displaystyle (e^{3}\cdot e^{9})=e^{12}$

When an exponent is raised to an exponent you multiply the exponents together. Hence, for $\displaystyle (e^{12})^{2}$ you would multiply $\displaystyle (12\cdot 2)$ to get $\displaystyle e^{24}$.

Answer: $\displaystyle e^{24}$

$\displaystyle 3^{2x+1}=81$

$\displaystyle 3^{2x+1}=3^4$

$\displaystyle \log_33^{2x+1}=\log_33^4$

$\displaystyle 2x+1=4$

$\displaystyle 2x=3$

$\displaystyle x=\frac{3}{2}$