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

  1. #1
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    conversion of log to the base of 4 to log to the base of 2

    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)
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  2. #2
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    If you use LaTex it is a lot easier to read.
    [tex]\log_{16}(x)[/tex] gives $\displaystyle \log_{16}(x)$.
    [tex]\frac{\log_{16}(x)}{\log_2(x)}[/tex] gives $\displaystyle \frac{\log_{16}(x)}{\log_2(x)}$.
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  3. #3
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    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?
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  4. #4
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    $\displaystyle \log_{16}(x)=\frac{\log(x)}{\log(16)}=\frac{\log(x )}{4\log(2)}=\frac{1}{4}\log_2(x)$.
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  5. #5
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    thank you so much.
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