Simplifying the exponential expression $e^{-4\ln x +8\ln y +2}$

Question

I'm totally stuck on this. Tried numerous sites for a decent explanation but can't find anything.

Simplify the expression $$e^{-4\ln x +8\ln y +2}.$$

Thanks in advance.

Answer 1

\begin{align} e^{-4\ln\left(x\right)+8\ln\left(y\right)+2}&=e^{-4\ln x}e^{8\ln y}e^2\\\\ &=e^2\frac{\left(e^{\ln y}\right)^8}{\left(e^{\ln x}\right)^4}\\\\ &=\displaystyle\boxed{\displaystyle\frac{e^2y^8}{x^4}} \end{align}

Answer 2

We may simplify this to \begin{align} f\left(x,y\right)=e^{-4\log\left|x\right|+8\log\left|y\right|+2}=\frac{e^{8\log\left|y\right|}e^{2}}{e^{4\log\left|x\right|}}=\frac{e^{\log\left|y^8\right|}e^2}{e^{\log\left|x^4\right|}}=\frac{e^2y^8}{x^4}, \tag{1} \end{align} but notice that I, by default, use $\log\left|x\right|$ to be the same as $\log_e\left|x\right|$ (or "$\ln$") instead of $\log_{10}\left|x\right|$. Whenever you're attempting to solve or simplify things like this your first goal should be to try and get rid of the exponents. In this case it was nice because the exponents could be split up and all had a common base of $e$. Whatever the common base is, you use a $\log$ of that base to cancel them and be left with what is in their exponents.