How do you condense $(2log7)/3$ ?

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Answer 1 · 2016-09-02T21:11:01

$(2log7)/3=log root3 49$

Here we may use the identities $mloga=loga^m$ and $1/nloga=log(a^(1/n))=log(root(n)a)$

Hence $(2log7)/3$

= $log7^2/3$

= $1/3 log49$

= $logroot3 49$

How do you condense (2log7)/3(2log7)/3?