Question

The value of the expression antilog $\left(\log _{\frac{1}{27}}\right.$ antilog $\left.\log _{\frac{1}{10}} 7\right)$ is in the form of $\sqrt[b]{a}$, where a, $b \in N$ then find the least value of $(a+b-2)$.

Answer 1

$\left(\log _{\frac{1}{27}} \operatorname{antilog}_{10} \log _{\frac{1}{10}} 7\right)=\operatorname{antilog}_3\left(\log _{\frac{1}{27}}(10)^{\log _{\frac{1}{10}}(7)}\right)=3^{\log _{27}(7)}=7^{1 / 3} $
$\Rightarrow \quad a =7, b =3 \Rightarrow \quad( a + b -2)=8$