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:
We need to evaluate a composite function involving logarithms. First, simplify the expression for \(h(x)\) using logarithm properties before substituting the complex \(x\) value.
Step 2: Key Formula or Approach:
Use the property of logarithms: \(\log a + \log b + \log c = \log(a \cdot b \cdot c)\).
Simplify \(h(x)\) fully, then substitute \(x = \frac{1}{3}e^{1/3}\).
Step 3: Detailed Explanation:
Construct \(h(x)\):
\[ h(x) = \log_e(9x) + \log_e(9x^2) + \log_e(9x^3) \] Combine using the product rule:
\[ h(x) = \log_e(9x \cdot 9x^2 \cdot 9x^3) \] \[ h(x) = \log_e(729x^6) \] Now, evaluate this for \(x = \frac{1}{3}e^{1/3}\).
First, find \(x^6\):
\[ x^6 = \left(\frac{1}{3}e^{1/3}\right)^6 = \left(\frac{1}{3}\right)^6 \cdot (e^{1/3})^6 \] \[ x^6 = \frac{1}{3^6} \cdot e^{6/3} = \frac{1}{729} \cdot e^2 \] Substitute \(x^6\) back into \(h(x)\):
\[ h\left(\frac{1}{3}e^{1/3}\right) = \log_e\left(729 \cdot \frac{1}{729} e^2\right) \] \[ = \log_e(e^2) \] Using the power rule of logarithms:
\[ = 2\log_e(e) = 2 \times 1 = 2 \] Step 4: Final Answer:
The value is 2.