# Thread: One question on logs.

1. ## One question on logs.

I need to know how to do this without a calculator.

http://img35.imageshack.us/img35/8409/logr.jpg

2. $\displaystyle - 2\log _a (3) = \log _a \left( {3^{ - 2} } \right) = \log _a \left( {\frac{1}{9}} \right)$

3. Thanks, but how will I know the answers to those questions w.o my calculator?

4. Originally Posted by NYCKid09
Thanks, but how will I know the answers to those questions w.o my calculator?
You have been shown why $\displaystyle \log _a \left( {\frac{1}{9}} \right) = - 2\log _a (3)$. And the question tells you the value of $\displaystyle \log_a (3)$. So what's the trouble ...?

5. log base a (3) = 0.5 is the starting equation. You want to find what log base a (1/9) is.

We can agree that one-ninth is simply 3^-2 (three to the power of negative two), for 1/3^2 would be 1/9. So now we have log base a (3^-2).

If you don't remember this from your logarithm rules, the exponent can be placed to the left of the log. So we go from log base a (3^-2) to [-2 * log base a (3)].

You are told that log base a (3) equals 0.5, so what's 0.5 * -2? Hopefully you know that this is -1 without needing a calculator.