# Log Law Question

1. Oct 14, 2009

### Senjai

1. The problem statement, all variables and given/known data
This is not a homework question.

Understanding how $$log_{\frac{1}{2}}\frac{1}{9} = log_2x^2$$

3. The attempt at a solution

Somehow, the base of the first logarith was turned in to 2^-1, no problem, but he was able to put the negative up in front of a log to equal:

$$-log_2\frac{1}{9}$$ How do you do that?

2. Oct 14, 2009

### Staff: Mentor

Let's look at the left side $$log_{\frac{1}{2}}\frac{1}{9}$$

What this log means is the exponent on 1/2 that produces 1/9. An equivalent equation is (1/2)y = 1/9. This is turn is equivalent to 1/(2y) = 1/9, or equivalently, 2y = 9.

Your equation can be rewritten as log2 9 = log2 x2, and from this we see that x2 = 9, which has two solutions.

3. Oct 14, 2009

### Senjai

Thanks! that.. makes things really easy to understand... If i don't know the proof I won't remember it.. lol. Thank you so much :) Test today.. :(

*edit*

just to make sure, the solution is +/- 3 correct?

4. Oct 14, 2009

Yes.