Does $e^{\ln(x^2)/\log(x)} = e^{x\ln(10)}$

Question

So this is the equation: $y=|e^{(\ln(x^2)/\log(x)}|$

We can convert log(x) to base e with the change of base rule: $\log(x) = \frac{\ln(x)}{\ln(10)}$

Then the expression $(\ln(x^2)/\log(x))$ equals $\frac{\ln(x^2)}{\ln(x)/\ln(10)}$, which is equal to \begin{align} \ln(x^2) \cdot \frac{\ln(10)}{\ln(x)} &= 2\ln(x) \cdot \frac{\ln(10)}{\ln(x)}\\ &= 2\ln(10) \end{align} So $e^{\ln(x^2)/\log(x)} = e^{2\ln(10)}$.

If you graph the former equation in the question is equals $y=100$, but when you graph the latter, you get some logarithmic curve.

EDIT

I figured the problem out now, in the 3rd line, I set $\ln(x^2)$ to $x\ln(x)$ as opposed to $2\ln(x)$

When is $e^x$ negative? Why have $||$? – Karolis Juodelė Dec 23 '13 at 20:05 — Karolis Juodelė, Dec 23 '13 at 20:05

score 2 · Accepted Answer · answered Dec 23 '13 at 20:04

2

$$\ln x^2 = 2 \ln x$$

Compare $e^{\ln x^2}$ with $e^{2 \ln x}$ if it's not clear.

answered Dec 23 '13 at 20:04

Karolis Juodelė

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OH I see my mistake now! Let me fix it. – user16795 Dec 23 '13 at 20:08

Does $e^{\ln(x^2)/\log(x)} = e^{x\ln(10)}$

1 Answers1