# Logarithmic

• May 11th 2008, 08:42 PM
aphan19
Logarithmic
I have a calc final tomorrow and I can't seem to figure this problem out. Please help. He gave us the answer which was 1.500, but I don't know how he got it.
Let logb A = 5 and logb B = -2. Find logb 2√AB.

• May 11th 2008, 08:44 PM
Mathstud28
Quote:

Originally Posted by aphan19
I have a calc final tomorrow and I can't seem to figure this problem out. Please help. He gave us the answer which was 1.500, but I don't know how he got it.
Let logb A = 5 and logb B = -2. Find logb 2√AB.

$\log_b(2\sqrt{AB})=\log_b(2)+\log_b(\sqrt{AB})=$ $\log_b(2)+\frac{1}{2}\log_b(AB)=\log_b(2)+\frac{1} {2}\bigg[\log_b(A)+\log_b(B)\bigg]$

Can you take it from there?
• May 11th 2008, 08:54 PM
aphan19
Quote:

Originally Posted by Mathstud28
$\log_b(2\sqrt{AB})=\log_b(2)+\log_b(\sqrt{AB})=$ $\log_b(2)+\frac{1}{2}\log_b(AB)=\log_b(2)+\frac{1} {2}\bigg[\log_b(A)+\log_b(B)\bigg]$

Can you take it from there?

Yes, I think I can take it from there, but did you get the 1/2 from the square root?
• May 11th 2008, 08:56 PM
o_O
$\sqrt{a} = a^{\frac{1}{2}}$
• May 11th 2008, 09:04 PM
aphan19
I'm stuck on logb (2). When I add it to the rest of the problem it came out to be a completely different answer.
[/COLOR][/FONT][/color]
• May 11th 2008, 09:08 PM
Mathstud28
Quote:

Originally Posted by aphan19
I'm stuck on logb (2). When I add it to the rest of the problem it came out to be a completely different answer.
[/color][/font][/color]

I dont think that two should be there

Because if it wasnt you would just ahve $\frac{1}{2}\bigg[\log_b(A)+\log_b(B)\bigg]=\frac{1}{2}\bigg[5+-2\bigg]=\frac{1}{2}\cdot{3}=\frac{3}{2}=1.5$