Question

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

Answer 1

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

Explanation:

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`