Discussion
Let \(f(x) = \log_{e}(9x)\) for \(x \gt 0\) and \(h(x) = f(x) + f(x^{2}) + f(x^{3})\). Then the value of \(h\left(\frac{1}{3}e^{\frac{1}{3}}\right)\) is equal to
Answer: B
The value is 2.
Simplify \(h\) with the rules of logarithms:
\[ \begin{aligned} h(x) &= \ln 9x + \ln 9x^{2} + \ln 9x^{3} \\ &= 3\ln 9 + 6\ln x \end{aligned} \]
With \(x = \frac{1}{3}e^{1/3}\), \(\ln x = -\ln 3 + \frac{1}{3}\):
\[ \begin{aligned} h &= 6\ln 3 + 6\left(-\ln 3 + \tfrac{1}{3}\right) \\ &= 6\ln 3 - 6\ln 3 + 2 = 2 \end{aligned} \]
Here \(3\ln 9 = 6\ln 3\), which is why the logarithms cancel completely.
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