Rewrite Log Function in terms of A and B?

**chopper07** · October 8th 2012, 01:02 PM

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.

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

A=Log₃4 , B=log₃5

this equation--> log₃(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.

**Plato** · October 8th 2012, 01:13 PM

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.

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

A=Log₃4 , B=log₃5

this equation--> log₃(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: $\log_3(15)=1+B$

**ebaines** · October 8th 2012, 01:20 PM

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

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

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

**chopper07** · October 8th 2012, 01:24 PM

Ohh, ok i get it now. thank you !!

**chopper07** · October 8th 2012, 01:27 PM

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

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