We are given the equation:

\[
0.2715 = \log(x)^2
\]

### Step 1: Take the square root of both sides

First, take the square root of both sides to eliminate the square on the logarithm:

\[
\sqrt{0.2715} = \log(x)
\]

\[
\log(x) = \pm \sqrt{0.2715}
\]

Calculating the square root of 0.2715:

\[
\log(x) = \pm 0.521
\]

### Step 2: Solve for \( x \)

Now, exponentiate both sides of the equation to solve for \( x \). Remember, \( \log(x) \) typically means logarithm to base 10, so we will raise 10 to the power of both sides:

\[
x = 10^{\pm 0.521}
\]

This gives two possible solutions:

\[
x = 10^{0.521} \quad \text{or} \quad x = 10^{-0.521}
\]

Calculating both values:

\[
x_1 = 10^{0.521} \approx 3.33
\]
\[
x_2 = 10^{-0.521} \approx 0.30
\]

### Final Answer:

\[
x \approx 3.33 \quad \text{or} \quad x \approx 0.30
\]

Would you like further details on any step?

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Here are 5 related questions you might find interesting:
1. What happens if we have \( \log_b(x) \) instead of \( \log(x) \)?
2. How do logarithmic properties help in solving exponential equations?
3. Can we solve \( \log(x)^3 = y \) similarly?
4. How would we solve \( \ln(x)^2 = z \)?
5. What are the key differences between natural logarithms and base-10 logarithms?

**Tip:** Logarithms and exponents are inverse operations—always remember this relationship when solving such equations!

Learn how to solve the logarithmic equation 0.2715 = log(x)^2 using square root and exponentiation methods. This detailed step-by-step solution covers key logarithmic properties and the inverse relationship between logarithms and exponents. Suitable for students in Grades 10-12.

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