<p>If <span class='math'>f(x) = \log_{10} x</span> and <span class='math'>g(x) = e^{0.1x}</span>, and if <span class='math'>h(x) = \frac{f(x^2) + g(x^2)}{(f(x^2))^{g(x^2)}} \cdot \frac{f(x^3) + g(x^3)}{(f(x^3))^{g(x^3)}}</span>, then the value of <span class='math'>h(10)</span> is</p>
Step-by-Step Solution
Key Concept: Recognize that when f(x) = log₁₀(x) and a function has the form [f(a) + g(a)]/[f(a)^g(a)], we need to check if the numerator equals the exponent base raised to its exponent, making the fraction equal to 1. This simplification is key to evaluating h(10).
<p><strong>Step 1: Define the given functions.</strong></p><p>f(x) = log₁₀(x) and g(x) = e^(0.1x)</p><p>h(x) = [f(x²) + g(x²)]/[f(x²)^g(x²)] · [f(x³) + g(x³)]/[f(x³)^g(x³)]</p><p><strong>Step 2: Calculate f(x²) and g(x²) when x = 10.</strong></p><p>f(10²) = f(100) = log₁₀(100) = log₁₀(10²) = 2</p><p>g(10²) = g(100) = e^(0.1 × 100) = e^10</p><p><strong>Step 3: Evaluate the first fraction at x = 10.</strong></p><p>First fraction = [f(100) + g(100)]/[f(100)^g(100)]</p><p>= [2 + e^10]/[2^(e^10)]</p><p><strong>Step 4: Calculate f(x³) and g(x³) when x = 10.</strong></p><p>f(10³) = f(1000) = log₁₀(1000) = log₁₀(10³) = 3</p><p>g(10³) = g(1000) = e^(0.1 × 1000) = e^100</p><p><strong>Step 5: Evaluate the second fraction at x = 10.</strong></p><p>Second fraction = [f(1000) + g(1000)]/[f(1000)^g(1000)]</p><p>= [3 + e^100]/[3^(e^100)]</p><p><strong>Step 6: Recognize the structural issue.</strong></p><p>Notice that neither fraction simplifies to 1 in the conventional sense. However, examining the problem structure reveals that for each term of the form [a + b]/[a^b] where a is small relative to b (and both grow exponentially), the expression tends toward 0 as b → ∞.</p><p><strong>Step 7: Alternative insight - Check if numerator could equal denominator base.</strong></p><p>Upon careful inspection, if the numerator equals the base of the exponent (i.e., f(x^n) + g(x^n) = f(x^n)), then the entire fraction becomes 1. But this doesn't hold here. Instead, the key observation is that each fraction has an exponentially large exponent in the denominator, making each fraction approach 0.</p><p><strong>Step 8: Product of two nearly-zero quantities.</strong></p><p>Since both fractions have exponentially large exponents in their denominators relative to their numerators, the product h(10) approaches 0.</p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C