# Logarithms

• Oct 12th 2008, 04:37 PM
lmschneider
Logarithms
14^log14 of 21
• Oct 12th 2008, 04:43 PM
Jhevon
Quote:

Originally Posted by lmschneider
14^log14 of 21

there is a rule: $\displaystyle a^{\log_a x} = x$
• Oct 12th 2008, 04:45 PM
lmschneider
Do you happen to know what that rule is called?
• Oct 12th 2008, 04:46 PM
Jhevon
Quote:

Originally Posted by lmschneider
Do you happen to know what that rule is called?

no, it is just one of the log rules that show up. it is very likely to be in a list of log rules in your text
• Oct 12th 2008, 04:48 PM
lmschneider
I looked and it's not, but that's okay. I will keep looking. So the answer is 21 then?
• Oct 12th 2008, 05:06 PM
Jhevon
Quote:

Originally Posted by lmschneider
I looked and it's not, but that's okay. I will keep looking. So the answer is 21 then?

yes

it is not hard to prove though. i will show you how to prove it. that way, if you are not allowed to use this rule, this is how you would attempt the problem in general.

Let $\displaystyle a^{\log_a x} = y$, we will show that $\displaystyle y = x$

$\displaystyle a^{\log_a x} = y$ ...........................................take $\displaystyle \log$ to the base $\displaystyle a$ of both sides

$\displaystyle \Rightarrow \log_a a^{\log_a x} = \log_a y$

$\displaystyle \Rightarrow \log_a x \cdot \log_a a = \log_a y$ .................since $\displaystyle \log_a (x^n) = n \log_a x$

$\displaystyle \Rightarrow \log_a x = \log_a y$ ...............................since $\displaystyle \log_a a = 1$

$\displaystyle \Rightarrow x = y$ ..............................................by taking the antilog of both sides.