Question

The value of $a^{\log_b c} - c^{\log_b a}$, where $a, b, c>0$ but $a, b, c \neq 1$, is

• A. a
• B. b
• C. c
• D. 0

Hint:

Remember that the identity \( a^{\log_b c} = c^{\log_b a} \) simplifies many logarithmic problems. It’s a useful property to recognize when dealing with logarithmic terms in exponents.

Answer 1

The Correct Option is D

To solve the problem, we need to prove that \( a^{\log_b c} - c^{\log_b a} = 0 \) using logarithmic identities.

1. Define the Initial Expression:
Let \( y = c \log_b a \). This represents a logarithmic relationship between the variables.

2. Take the Logarithm of Both Sides:
Taking the logarithm (base \( c \)) on both sides:

\( \log_c y = \log_b a \).

3. Apply the Change of Base Formula:
Using the logarithmic identity \( \log_k m = \frac{\log m}{\log k} \), we convert both sides to natural logarithms (or base-10):

\( \frac{\log y}{\log c} = \frac{\log a}{\log b} \).

4. Rearrange the Equation:
Cross-multiply to isolate \( \log y \):

\( \frac{\log y}{\log a} = \frac{\log c}{\log b} \).

5. Rewrite in Logarithmic Form:
This implies \( \log_a y = \log_b c \), which means \( y = a^{\log_b c} \).

6. Substitute Back the Original Expression:
Since \( y = c \log_b a \), we substitute to get:

\( c \log_b a = a^{\log_b c} \).

7. Prove the Given Identity:
By symmetry, \( c^{\log_b a} = a^{\log_b c} \). 

Therefore: \( a^{\log_b c} - c^{\log_b a} = 0 \).

Final Answer:
The proof shows that \( \boxed{a^{\log_b c} - c^{\log_b a} = 0} \).

Answer 2

We are given the expression \( a^{\log_b c} - c^{\log_b a} \). 
Let's simplify it step by step. 
First, recall the logarithmic identity that \( a^{\log_b c} = c^{\log_b a} \), which is a general property of logarithms. Therefore, the given expression becomes: \[ a^{\log_b c} - c^{\log_b a} = 0. \] 
Thus, the value of the given expression is 0.