To solve the equation $\log(6^x) = 2$, follow these steps:

1. Use the logarithmic property $\log(a^b) = b \cdot \log(a)$: $x \cdot \log(6) = 2$

2. Solve for $x$ by dividing both sides by $\log(6)$: $x = \frac{2}{\log(6)}$

3. Use a calculator to evaluate $\log(6)$. Assuming a base-10 logarithm: $\log(6) \approx 0.77815$

4. Compute $x$: $x = \frac{2}{0.77815} \approx 2.57$

Final Answer:

$x \approx 2.57$

Would you like further clarification or a deeper explanation?

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Related Questions:

1. How do properties of logarithms simplify solving exponential equations?
2. How does changing the base of a logarithm affect the calculation?
3. What is the difference between natural logarithms ($\ln$) and common logarithms ($\log$)?
4. Can logarithms be applied to solve equations with fractional exponents?
5. How do you solve equations involving $e^x$ using natural logarithms?

Tip:

Always verify the logarithm base being used ($\log$ for base-10 or $\ln$ for base-( e