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

Jan 2010
26
1
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
Aug 2006
22,460
8,632
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 \(\displaystyle \frac{\log_{16}(x)}{\log_2(x)}\).
 
Jan 2010
26
1
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
Aug 2006
22,460
8,632
\(\displaystyle \log_{16}(x)=\frac{\log(x)}{\log(16)}=\frac{\log(x)}{4\log(2)}=\frac{1}{4}\log_2(x)\).