1. ## 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.

2. 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.

$\displaystyle \log_b(2\sqrt{AB})=\log_b(2)+\log_b(\sqrt{AB})=$$\displaystyle \log_b(2)+\frac{1}{2}\log_b(AB)=\log_b(2)+\frac{1} {2}\bigg[\log_b(A)+\log_b(B)\bigg] Can you take it from there? 3. Originally Posted by Mathstud28 \displaystyle \log_b(2\sqrt{AB})=\log_b(2)+\log_b(\sqrt{AB})=$$\displaystyle \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?

4. $\displaystyle \sqrt{a} = a^{\frac{1}{2}}$

5. 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.
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6. 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.
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I dont think that two should be there

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