Brock Byrd

2022-07-01

${e}^{\mathrm{ln}((9)/\mathrm{log}(3))}=100$? Why is that?

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

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

escampetaq5

Beginner2022-07-02Added 12 answers

In the calculator $\mathrm{log}$ is for the decimal logarithm and $\mathrm{ln}$ for the natural logarithm so

$\frac{\mathrm{ln}9}{\mathrm{log}3}=\frac{2\mathrm{ln}3\times \mathrm{ln}10}{\mathrm{ln}3}=\mathrm{ln}100$

$\frac{\mathrm{ln}9}{\mathrm{log}3}=\frac{2\mathrm{ln}3\times \mathrm{ln}10}{\mathrm{ln}3}=\mathrm{ln}100$

lilmoore11p8

Beginner2022-07-03Added 6 answers

We can convert $\mathrm{log}(3)$ to base $e$ with the change of base rule:

$\mathrm{log}(3)=\frac{\mathrm{ln}(3)}{\mathrm{ln}(10)}$

Then the expression $(\mathrm{ln}(9)/\mathrm{log}(3))$ equals $\frac{ln(9)}{\mathrm{ln}(3)/\mathrm{ln}(10)}$, which is equal to

$\begin{array}{rl}\mathrm{ln}(9)\cdot \frac{\mathrm{ln}(10)}{\mathrm{ln}(3)}& =2\mathrm{ln}(3)\cdot \frac{\mathrm{ln}(10)}{\mathrm{ln}(3)}\\ & =2\cdot \mathrm{ln}(10)\\ & =\mathrm{ln}(100)\end{array}$

So ${e}^{\mathrm{ln}(9)/\mathrm{log}(3)}={e}^{\mathrm{ln}(100)}$. We know ${e}^{\mathrm{ln}(a)}=a$, so ${e}^{\mathrm{ln}(100)}=100$

$\mathrm{log}(3)=\frac{\mathrm{ln}(3)}{\mathrm{ln}(10)}$

Then the expression $(\mathrm{ln}(9)/\mathrm{log}(3))$ equals $\frac{ln(9)}{\mathrm{ln}(3)/\mathrm{ln}(10)}$, which is equal to

$\begin{array}{rl}\mathrm{ln}(9)\cdot \frac{\mathrm{ln}(10)}{\mathrm{ln}(3)}& =2\mathrm{ln}(3)\cdot \frac{\mathrm{ln}(10)}{\mathrm{ln}(3)}\\ & =2\cdot \mathrm{ln}(10)\\ & =\mathrm{ln}(100)\end{array}$

So ${e}^{\mathrm{ln}(9)/\mathrm{log}(3)}={e}^{\mathrm{ln}(100)}$. We know ${e}^{\mathrm{ln}(a)}=a$, so ${e}^{\mathrm{ln}(100)}=100$

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