Log Law: Understanding -log_2\frac{1}{9}

[Senjai]

Homework Statement

This is not a homework question.

Understanding how [tex] log_{\frac{1}{2}}\frac{1}{9} = log_2x^2 [/tex]

The Attempt at a Solution

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

[tex] -log_2\frac{1}{9} [/tex] How do you do that?

 

[Mark44]

Let's look at the left side [tex] log_{\frac{1}{2}}\frac{1}{9} [/tex]

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/(2^y) = 1/9, or equivalently, 2^y = 9.

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

 

[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?

 

[Mark44]

Yes.