Rewrite Log Function in terms of A and B?

This has been driving me up the wall, I have a test the day after tomorrow and this is DEFINITELY going to be on the test. It seems simple, but for some reason I can't get the hang of it, please help. (Crying)

***Rewrite the equation in terms of A and B

A=Log_{3}4 , B=log_{3}5

this equation--> log_{3}(15/2)

I've tried 3b+(a/2), but thats wrong. I know why it's wrong, but I dont know how to get the right answer.

Re: Rewrite Log Function in terms of A and B?

Quote:

Originally Posted by

**chopper07** This has been driving me up the wall, I have a test the day after tomorrow and this is DEFINITELY going to be on the test. It seems simple, but for some reason I can't get the hang of it, please help. (Crying)

***Rewrite the equation in terms of A and B

A=Log_{3}4 , B=log_{3}5

this equation--> log_{3}(15/2)

I've tried 3b+(a/2), but thats wrong. I know why it's wrong, but I dont know how to get the right answer.

Hint: $\displaystyle \log_3(15)=1+B$

Re: Rewrite Log Function in terms of A and B?

$\displaystyle A = \log_3(4) = \log_3(2^2) = 2 log_3(2)$, so$\displaystyle \log_3(2) = \frac A 2$.

$\displaystyle \log_3(15/2) = \log_3(\frac {3*5} 2 ) = \log_3(3) + \log_3 (5) - \log_3 (2)$

You have values for $\displaystyle \log_3(5)$ and $\displaystyle \log_3 (2)$. The only thing left is $\displaystyle \log_3(3)$. Recall that if $\displaystyle \log_a b = c$ then $\displaystyle a^c = b$. Here a = 3 and b = 3, so what's the value of c? Hence what's the value of $\displaystyle \log_3 (3)$? Post back with your solution and we'll check it for you.

Re: Rewrite Log Function in terms of A and B?

Ohh, ok i get it now. thank you !! (Hi)

Re: Rewrite Log Function in terms of A and B?

Wait, so would it be 1+B+(a/2) ?