# Dazed and Confused

• Apr 20th 2012, 04:42 PM
Past45
Dazed and Confused
If log b 3=0.5
log b 5= 0.7

evaluate log b √3
√6

Log b √3=logb 3^1/2

=1/2 logb 3

1/2(0.5)
=.25

Am I on the right track? And if I am, the √6 is somewhat throwing me off. Can I get help with that please? Ty
• Apr 20th 2012, 05:07 PM
Plato
Re: Dazed and Confused
Quote:

Originally Posted by Past45
If log b 3=0.5
log b 5= 0.7

evaluate log b √3
√6

Log b √3=logb 3^1/2

=1/2 logb 3

1/2(0.5)
=.25

Am I on the right track? And if I am, the √6 is somewhat throwing me off. Can I get help with that please? Ty

It is almost impossible to read what you posted.
Learn to use basic LaTeX.
[TEX]\log_b\sqrt{3}=\frac{\log_b(3)}{2}[/TEX] gives $\displaystyle \log_b\sqrt{3}=\frac{\log_b(3)}{2}$.
• Apr 20th 2012, 06:35 PM
Past45
Re: Dazed and Confused
Now, I am confused in what you wrote, lol.

So logb √3 equals to logb(3)/2?

Let me try to re-write this problem. No clue what this LaTeX thing is, so please forgive me.

If log b 3=0.5
logb 5= 0.7

Now I have to evaluate log √3 and log √6. My steps for √3 is already posted above this for that problem.
• Apr 21st 2012, 12:13 PM
HallsofIvy
Re: Dazed and Confused
If, by "log b" you mean "log base b", $\displaystyle log_b$ or log_b then

Use the laws of logarithms: $\displaystyle log_b(\sqrt{3})= log_b(3^{1/2})= (1/2)log_b(3)$ exactly as you say.

$\displaystyle log_b(\sqrt{6})= log_b(6^{1/2})= (1/2)log_b(6)= (1/2)log_b(2(3))$$\displaystyle = (1/2)(log_b(2)+ log_b(3))$.