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Note: I am using log base 10 and I am trying to rewrite the equation using exponents instead of logs.

Here is what I have and I am wondering if I did it correctly (if not how am I suspose to solve this question):

$$\log(A^2) = B$$ $$2\log(A) = \log(B)$$ $$\log(B)/\log(A) = 2$$ $$10^2$$

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  • $\begingroup$ Your question is not clear. What does the out of the blue $10^2$ mean? Are you trying to find $A$ in terms of $B$? Or something else? $\endgroup$ – Aryabhata Apr 19 '13 at 0:27
  • $\begingroup$ Are you trying to solve explicitly for $A$ and $B$? Also, from the first line to the second, you make a mistake. The $\log(B)$ should only be $B$. $\endgroup$ – Clayton Apr 19 '13 at 0:28
  • $\begingroup$ The instructions are: Rewrite each of the following using exponents instead of logs. The question is Log(A^2)=B. About the 10^2 I got that because for example log(0.01) <=> 10^-2 $\endgroup$ – Bob Apr 19 '13 at 0:33
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Your second line doesn't look correct. The rule is $\log(A^n)=n\cdot\log(A)$, so:

$$ \log(A^2)=B \\ 2\log(A)=B \\ \log(A)=\frac{1}{2}B \\ A=10^{\frac{1}{2}B} $$

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  • $\begingroup$ I've removed my down vote. Your solution looks correct now. $\endgroup$ – Clayton Apr 19 '13 at 0:31
  • $\begingroup$ Thank you - just a silly mistake! $\endgroup$ – Warren Moore Apr 19 '13 at 0:32

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