$\displaystyle log{\sqrt{3}} $ 9

Printable View

- Nov 26th 2007, 06:05 PMDarkhrse99Can i get help with this logerithm problem
$\displaystyle log{\sqrt{3}} $ 9

- Nov 26th 2007, 06:13 PMJhevon
- Nov 26th 2007, 06:23 PMDarkhrse99
I'm suppose to find the exact value of this logerithm. I believe that is the base.

- Nov 26th 2007, 06:36 PMJhevon
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 PMDarkhrse99
I'm guessing it's 2, since we have to raise 2 to$\displaystyle \sqrt{3} $ to equal 9.

- Nov 26th 2007, 07:11 PMJhevon
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 PMDarkhrse99
$\displaystyle \sqrt {3}^4=9 $ so the answer is 4.

- Nov 26th 2007, 07:38 PMJhevon
- Nov 26th 2007, 07:43 PMDarkhrse99
Cool!!