syifa01 syifa01

21-08-2016
Mathematics

Answered

2/3×2log64-1/2×2log16

Verified Answer :

Anonymous Anonymous

21-08-2016

\(\dfrac{2}{3} *2log 64 - \dfrac{1}{2} *2log 16\)

\(\dfrac{2}{3} *2log \ 2^6 - \dfrac{1}{3} *2 \ log \ 2^4\)

\(\dfrac{2}{3} *(6)*2log2 - \dfrac{1}{2} *(4)*2log2\)

\(4*(1) - 2*(1)\)

\(= 2\)

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