1. **State the problem:** Prove that $a^{\log_a x} = x$ for $a > 0$, $a \neq 1$, and $x > 0$. 2. **Recall the definition of logarithm:** The logarithm $\log_a x$ is defined as the exponent to which the base $a$ must be raised to get $x$. In other words, if $y = \log_a x$, then by definition $a^y = x$. 3. **Apply the definition:** Substitute $y = \log_a x$ into the expression $a^{\log_a x}$: $$a^{\log_a x} = a^y$$ 4. **Use the definition of $y$:** Since $a^y = x$, it follows that: $$a^{\log_a x} = x$$ 5. **Explanation:** This shows that raising $a$ to the power of the logarithm base $a$ of $x$ returns $x$ itself. This is because logarithms and exponentials with the same base are inverse functions. 6. **Graphical confirmation:** The graph of $y = a^{\log_a x}$ overlaps the line $y = x$ for $x > 0$, confirming the equality visually. **Final answer:** $$a^{\log_a x} = x$$