The value of $16^{\log_4 5}$ is

### Concept & Formula This problem tests the fundamental property of exponents involving logarithms. When a base is raised to a logarithm with the same base, they cancel each other out. $$ a^{\log_a x} = x $$ ### Step-by-Step Solution * **Given:** An exponential expression where the base is 16 and the exponent is a logarithm with base 4. $$ 16^{\log_4 5} $$ * **Calculation:** To apply the cancellation property, the main base (16) must match the logarithm's base (4). Rewrite 16 as a power of 4. $$ 16 = 4^2 $$ * Substitute this back into the original expression. $$ (4^2)^{\log_4 5} $$ * Apply the power of a power rule for exponents: $(x^a)^b = x^{ab}$. Multiply the exponents. $$ 4^{2 \times \log_4 5} $$ * Use the logarithmic power rule ($n \log_b x = \log_b (x^n)$) to move the coefficient 2 inside the logarithm as an exponent for 5. $$ 4^{\log_4 (5^2)} $$ $$ 4^{\log_4 25} $$ * Now the main base (4) perfectly matches the logarithm's base (4). Apply the fundamental identity $a^{\log_a x} = x$. $$ 25 $$ ### Exam Strategy & Shortcut An even faster shortcut relies on the swap property of exponential logarithms: $a^{\log_b c} = c^{\log_b a}$. Swap the 16 and the 5: $$ 16^{\log_4 5} = 5^{\log_4 16} $$ Since $\log_4 16 = 2$, the expression simply becomes $5^2 = 25$. ### Common Pitfall A frequent error is trying to apply the identity $a^{\log_a x} = x$ prematurely without matching the bases, mistakenly answering 5 because they ignore that 16 is not 4. Always ensure bases are identical before canceling. ### Final Answer **Therefore, the correct answer is 25.**