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#