# Can i get help with this logerithm problem

• Nov 26th 2007, 06:05 PM
Darkhrse99
Can i get help with this logerithm problem
$\displaystyle log{\sqrt{3}}$ 9
• Nov 26th 2007, 06:13 PM
Jhevon
Quote:

Originally Posted by Darkhrse99
$\displaystyle log{\sqrt{3}}$ 9

what exactly are you supposed to do with this? is $\displaystyle \sqrt{3}$ the base?
• Nov 26th 2007, 06:23 PM
Darkhrse99
I'm suppose to find the exact value of this logerithm. I believe that is the base.
• Nov 26th 2007, 06:36 PM
Jhevon
Quote:

Originally Posted by Darkhrse99
I'm suppose to find the exact value of this logerithm. I believe that is the base.

does it look like $\displaystyle \log_{\sqrt{3}}9$? if so, it is the base.

remember what a logarithm is, logarithms are powers.

Definition: the logarithm of a number to a given base is the power to which the base must be raised to give the number.

that is, if $\displaystyle \log_a b = c$ then $\displaystyle a^c = b$

example, $\displaystyle \log_28 = 3$ since we have to raise the base 2 to the power 3 to get 8

now, $\displaystyle \log_{\sqrt{3}}9$ is the power we have to raise $\displaystyle \sqrt{3}$ to to get 9. so what number is that?
• Nov 26th 2007, 06:42 PM
Darkhrse99
I'm guessing it's 2, since we have to raise 2 to$\displaystyle \sqrt{3}$ to equal 9.
• Nov 26th 2007, 07:11 PM
Jhevon
Quote:

Originally Posted by Darkhrse99
I'm guessing it's 2, since we have to raise 2 to$\displaystyle \sqrt{3}$ to equal 9.

what?! are you confused about something? where did 2 come from?

first of all, $\displaystyle 2^{\sqrt{3}}$ is not 9

second of all, there are no 2's here

third of all, we want to know what power we have to raise the base to, that is we want to solve $\displaystyle \sqrt{3}^x = 9$ for $\displaystyle x$
• Nov 26th 2007, 07:36 PM
Darkhrse99
$\displaystyle \sqrt {3}^4=9$ so the answer is 4.
• Nov 26th 2007, 07:38 PM
Jhevon
Quote:

Originally Posted by Darkhrse99
$\displaystyle \sqrt {3}^4=9$ so the answer is 4.

yes
• Nov 26th 2007, 07:43 PM
Darkhrse99
Cool!!