🧩 Concept Explanation

When two numbers \(a\) and \(b\) are raised to powers involving logarithms with the same base \(x\), they satisfy a useful identity that lets you swap the base and the argument.

🧮 Formula(s)

\[ a^{\log_x b} \;=\; b^{\log_x a} \]

Proof idea. Let \( \log_x a=p \) and \( \log_x b=q \Rightarrow a=x^{p},\, b=x^{q}\). Then \( a^{\log_x b}=(x^{p})^{q}=x^{pq}\) and \( b^{\log_x a}=(x^{q})^{p}=x^{qp}=x^{pq}\), hence equal.

❓ Example Question

Find the value of \(4^{\log_{16} 25}\).

✅ Solution / Shortcut

Using \(a^{\log_x b}=b^{\log_x a}\):

\[ 4^{\log_{16} 25} \;=\; 25^{\log_{16} 4}. \]

Now \( \log_{16}4=\log_{16}\!\left(16^{1/2}\right)=\tfrac{1}{2}\). Hence,

\[ 4^{\log_{16}25}=25^{1/2}=\sqrt{25}=5. \]

📝 Quick Quiz

Evaluate: \(27^{\log_{3} 2}\)