How do you simplify e^[log_e^2 (9)]?

Apr 7, 2018

The most simplified I could get it is ${81}^{\ln} 3$ which is about $124.935$.

Explanation:

Use these logarithmic properties:

$\textcolor{b l u e}{{\log}_{e}} x = \textcolor{b l u e}{\ln} x$

${\log}^{2} x = {\left(\log x\right)}^{2}$

$\log \left({\textcolor{b l u e}{x}}^{\textcolor{red}{a}}\right) = \textcolor{red}{a} \log \textcolor{b l u e}{x}$

And these exponential properties:

${\left({x}^{\textcolor{red}{m}}\right)}^{\textcolor{b l u e}{n}} = {x}^{\textcolor{red}{m} \textcolor{b l u e}{n}} = {\left({x}^{\textcolor{b l u e}{n}}\right)}^{\textcolor{red}{m}}$

${e}^{\ln} \textcolor{b l u e}{x} = \textcolor{b l u e}{x}$

Here's the expression:

$\textcolor{w h i t e}{=} {e}^{{\log}_{e}^{2} 9}$

$= {e}^{{\ln}^{2} \left(9\right)}$

$= {e}^{{\ln}^{2} 9}$

$= {e}^{{\left(\ln 9\right)}^{2}}$

$= {e}^{{\left(\ln \left({3}^{2}\right)\right)}^{2}}$

$= {e}^{{\left(2 \ln 3\right)}^{2}}$

$= {e}^{{2}^{2} {\ln}^{2} 3}$

$= {e}^{4 {\ln}^{2} 3}$

$= {\left({e}^{{\ln}^{2} 3}\right)}^{4}$

$= {\left({e}^{\ln 3 \cdot \ln 3}\right)}^{4}$

$= {\left({\left({e}^{\ln 3}\right)}^{\ln} 3\right)}^{4}$

$= {\left({\left({\textcolor{red}{\cancel{\textcolor{b l a c k}{e}}}}^{\textcolor{red}{\cancel{\textcolor{b l a c k}{\mathcal{\ln}}}} 3}\right)}^{\ln} 3\right)}^{4}$

$= {\left({3}^{\ln} 3\right)}^{4}$

$= {3}^{4 \ln 3}$

$= {\left({3}^{4}\right)}^{\ln} 3$

$= {81}^{\ln} 3$

$\approx 124.93528000 \ldots$

Hope this helped!