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If a,b,c are non-zero and $$14^a=36^b=84^c$$, then $$6b(\frac{1}{c}-\frac{1}{a})$$ is equal to
Correct Answer: 3
Let $$14^a=36^b=84^c$$ = k
=> a = $$\log_{14}k$$ , b = $$\log_{36}k$$, c=$$\log_{84}k$$
$$6b(\frac{1}{c}-\frac{1}{a})$$ = $$6\cdot\frac{1}{2}\log_6k\left(\log_k84-\log_k14\right)$$ = 3
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