How do you solve $log 8 - (1/3) log x = log 2$ ?

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Answer 1 · 2016-03-09T20:52:09

$x=64$

From the laws of logs we may write this as

$log(8/x^(1/3))=log2$

Now taking the antilog on both sides and cross-multiplying yields

$2x^(1/3)=8$

Now applying laws of exponents and surds and simplifying, we get

$root(3)x=4$

$thereforex=4^3=64$

How do you solve log 8 - (1/3) log x = log 2log 8 - (1/3) log x = log 2?