Question

1/2 log 9+2 log 6-1/4 log 81-log 12=3log3 prove

Answer 1

Step-by-step explanation:

$\implies$$log9^{\frac{1}{2}}+log6^{2}-log81^{\frac{1}{4}}-log12=log3$

$\implies$$log(3^{2})^{\frac{1}{2}}+log36-log(3^{4})^{\frac{1}{4}}-log12=log3$

$\implies$$(log3+log36)-(log3+log12)=log3$

$\implies$$(log3×36)-(log3×12)=log3$

$\implies$$log108-log36=log3$

$\implies$ $log{\Large\frac{108}{36}}=log3$

$\implies$ $log3=log3$

Question is wrong : RHS should be only log3.

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