Question

Let $f(x) = \log_e(9x)$ for $x > 0$ and $h(x) = f(x) + f(x^2) + f(x^3)$. Then the value of $h(\frac{1}{3}e^{1/3})$ is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Always combine logarithms into a single term before substituting complicated values. It often leads to major cancellations, as $729$ and $(1/3)^6$ did here.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem uses logarithmic properties: $\log A + \log B = \log(AB)$ and $\log(x^n) = n \log x$.

Step 2: Detailed Explanation:
1. Express $h(x)$ using logarithmic properties:
\[ h(x) = \log_e(9x) + \log_e(9x^2) + \log_e(9x^3) \]
\[ h(x) = \log_e(9x \cdot 9x^2 \cdot 9x^3) = \log_e(9^3 \cdot x^{1+2+3}) = \log_e(729x^6) \]
2. Substitute $x = \frac{1}{3}e^{1/3}$:
\[ x^6 = \left( \frac{1}{3}e^{1/3} \right)^6 = \left( \frac{1}{3} \right)^6 \cdot (e^{1/3})^6 = \frac{1}{729} \cdot e^2 \]
3. Substitute $x^6$ into the expression for $h(x)$:
\[ h = \log_e \left( 729 \cdot \frac{1}{729} \cdot e^2 \right) \]
\[ h = \log_e(e^2) = 2 \log_e(e) = 2(1) = 2 \]

Step 3: Final Answer:
The value of $h(\frac{1}{3}e^{1/3})$ is 2.