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

#### hangainlover

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

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

#### hangainlover

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

#### Plato

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

#### hangainlover

thank you so much.