Solve $\left(6^x\right)\left(3^{2x+1}\right)=4^{x+2}$, expressing the answer as the quotient of natural logs.

Question

This question has been bugging me for a while. While I can solve for $x$, I can't seem to express it as $x=\frac{\ln a}{\ln b}$ as required.

Without further ado here is the question — solve $$\left(6^x\right)\left(3^{2x+1}\right)=4^{x+2}$$ giving your answer in the form $$x=\frac{\ln a}{\ln b}$$ where $a$ and $b$ are rational numbers.

I tried many ways of solving this, eventually getting to $\frac{3x+1}{x+4}=\frac{\ln 2}{\ln 3}$. This didn't lead anywhere.

I'd appreciate some help!

Answer 1

I tried many ways of solving this, eventually getting to $\frac{3x+1}{x+4}=\frac{\ln 2}{\ln 3}$. This didn't lead anywhere.

This is fine. Did you solve this expression for $x$?

$$\frac{3x+1}{x+4}=\frac{\ln 2}{\ln 3} \iff x = \frac{\ln 3 -4\ln 2}{\ln 2 - 3\ln 3} = \ldots$$

Now use properties of logarithms to rewrite this expression towards the desired form.

Answer 2

Your last line is ok, then multiplying out we get $$3x\ln(3)+\ln(3)=x\ln(2)+4\ln(2)$$ and now combining like terms: $$x(3\ln(3)-\ln(2))=4\ln(2)-\ln(3)$$ Can you finish?