Question

pls answer this question

Answer 1

2a=3b=6c=k suppose
log2(base)k=a........ log

Answer 2

$Let \: \: 2 ^{a} = 3 ^{b} = 6 ^{c} = k \\2 = k ^{ \frac{1}{a} } \: 3 = k ^{ \frac{1}{b} } = 6 = k ^{ \frac{1}{c} } \\ 6 = 2 \times 3 \\ k ^{ \frac{1}{a} } \times k ^{ \frac{1}{b} } = k ^{ \frac{1}{c} } \\ k^{ \frac{1}{a} + \frac{1}{b} } = k ^{ \frac{1}{c} } \\ k^{ \frac{b + a}{ab} } = k ^{ \frac{1}{c} } \\ \frac{a + b}{ab} = \frac{1}{c} \\ c = \frac{ab}{a + b} \:$