14^log14 of 21

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- Oct 12th 2008, 04:37 PMlmschneiderLogarithms
14^log14 of 21

- Oct 12th 2008, 04:43 PMJhevon
- Oct 12th 2008, 04:45 PMlmschneider
Do you happen to know what that rule is called?

- Oct 12th 2008, 04:46 PMJhevon
- Oct 12th 2008, 04:48 PMlmschneider
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 PMJhevon
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.