Question

If $60^a=3$ and $60^b=5$, then $12^{\frac{(1-a-b)}{2(1-b)}}$ is equal to

• A. $\frac{1}{2}$
• B. 2
• C. 15
• D. $\frac{1}{15}$

Answer 1

Answer: (B)

$(60{)}^a=3\Rightarrow a=\frac{\log 3}{\log 60};(60{)}^b=5\Rightarrow b=\frac{\log 5}{\log 60}$
Now, $\frac{1-a-b}{2(1-b)}=\frac{\left(1-\frac{\log 3}{\log 60}-\frac{\log 5}{\log 60}\right)}{2\left(1-\frac{\log 5}{\log 60}\right)}=\frac{(\log 60-\log 15)}{2(\log 60-\log 5)}=\frac{\log 4}{2\log 12}={\log }_{12}(2)$
$\therefore 12^{\frac{(1-a-b)}{2(a-b)}}=12^{{\log }_{12}(2)}=2$