Correct Answer - Option 2 : 100

**Concept:**

- \({\log _a}b = \frac{{\log b}}{{\log a}}\)

**Calculation:**

Given: \(\frac{{\log 256}}{{\log 16}} = \log x\)

\(⇒ \frac{{\log {{\left( {16} \right)}^2}}}{{\log \left( {16} \right)}} = \log x\)

As we know that, \({\log _a}b = \frac{{\log b}}{{\log a}}\) and \({\log _a}{a^n} = n\) where a ≠ 1

⇒ 2 = log x

Raising both sides to the power 10, we get

⇒ x = 10^{2}

⇒ x = 100