Brock Byrd

2022-07-01

${e}^{\mathrm{ln}\left(\left(9\right)/\mathrm{log}\left(3\right)\right)}=100$? Why is that?
I was messing around with my calculator and meant to put in $\mathrm{log}\left(9\right)/\mathrm{log}\left(3\right)$, which was supposed to be two. I look back, and I see my mistake, but curiously, I input that to the function ${e}^{\left(x\right)}$ and it comes out with $100$... Why is that? Why is ${e}^{\mathrm{ln}\left(9\right)/\mathrm{log}\left(3\right)}=100$?

escampetaq5

In the calculator $\mathrm{log}$ is for the decimal logarithm and $\mathrm{ln}$ for the natural logarithm so
$\frac{\mathrm{ln}9}{\mathrm{log}3}=\frac{2\mathrm{ln}3×\mathrm{ln}10}{\mathrm{ln}3}=\mathrm{ln}100$

lilmoore11p8

We can convert $\mathrm{log}\left(3\right)$ to base $e$ with the change of base rule:
$\mathrm{log}\left(3\right)=\frac{\mathrm{ln}\left(3\right)}{\mathrm{ln}\left(10\right)}$
Then the expression $\left(\mathrm{ln}\left(9\right)/\mathrm{log}\left(3\right)\right)$ equals $\frac{ln\left(9\right)}{\mathrm{ln}\left(3\right)/\mathrm{ln}\left(10\right)}$, which is equal to
$\begin{array}{rl}\mathrm{ln}\left(9\right)\cdot \frac{\mathrm{ln}\left(10\right)}{\mathrm{ln}\left(3\right)}& =2\mathrm{ln}\left(3\right)\cdot \frac{\mathrm{ln}\left(10\right)}{\mathrm{ln}\left(3\right)}\\ & =2\cdot \mathrm{ln}\left(10\right)\\ & =\mathrm{ln}\left(100\right)\end{array}$
So ${e}^{\mathrm{ln}\left(9\right)/\mathrm{log}\left(3\right)}={e}^{\mathrm{ln}\left(100\right)}$. We know ${e}^{\mathrm{ln}\left(a\right)}=a$, so ${e}^{\mathrm{ln}\left(100\right)}=100$

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