conversion of log to the base of 4 to log to the base of 2

**hangainlover** · May 26th 2010, 12:42 PM

i know that log to the base of a of b can be re-written as (log of b)/(log of a)
But, i got lost when i saw a question
log to the base of 16 (x) + log to the base of 4 (x) + log to the base of 2 (x) = 7

how would you go about solving this question?
I thought about re-defining every single term of those logs in to (log of b)/(log of a) shape, but it doesnt give me a whole number answer (answer is 16)

**Plato** · May 26th 2010, 12:53 PM

If you use LaTex it is a lot easier to read.
[tex]\log_{16}(x)[/tex] gives $\log_{16}(x)$ .
[tex]\frac{\log_{16}(x)}{\log_2(x)}[/tex] gives $\frac{\log_{16}(x)}{\log_2(x)}$ .

**hangainlover** · May 26th 2010, 12:58 PM

Thanks for the info, i will rewrite it to make it more clear
$\log_{16}(x)$ + $\log_{4}(x)$ + $\log_{2}(x)$ = 7
i couldnt find x (the hint in the book says i should be able to change $\log_{16}(x)$ and $\log_{4}(x)$ to the $\log_{2}(x)$ form
but i cant do that.
how should i make that happen?

**Plato** · May 26th 2010, 01:30 PM

$\log_{16}(x)=\frac{\log(x)}{\log(16)}=\frac{\log(x )}{4\log(2)}=\frac{1}{4}\log_2(x)$ .

**hangainlover** · May 26th 2010, 01:36 PM

thank you so much.

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