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: \(\displaystyle e^{24}\)

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

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

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

\(\displaystyle 2x+1=4\)

\(\displaystyle 2x=3\)

\(\displaystyle x=\frac{3}{2}\)