# simple logarithms with exponents

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):

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

• 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? – Aryabhata Apr 19 '13 at 0:27
• 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 $B$. – Clayton Apr 19 '13 at 0:28
• 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 – Bob Apr 19 '13 at 0:33

Your second line doesn't look correct. The rule is $\log(A^n)=n\cdot\log(A)$, so:
$$\log(A^2)=B \\ 2\log(A)=B \\ \log(A)=\frac{1}{2}B \\ A=10^{\frac{1}{2}B}$$