Logarithms

**lmschneider** · October 12th 2008, 04:37 PM

14^log14 of 21

**Jhevon** · October 12th 2008, 04:43 PM

Originally Posted by lmschneider

14^log14 of 21

there is a rule: $a^{\log_a x} = x$

**lmschneider** · October 12th 2008, 04:45 PM

Do you happen to know what that rule is called?

**Jhevon** · October 12th 2008, 04:46 PM

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

**lmschneider** · October 12th 2008, 04:48 PM

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

**Jhevon** · October 12th 2008, 05:06 PM

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

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

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

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

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

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

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