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The equation you've provided is:

\[
\log\left(\frac{r}{r_2}\right) = -\frac{6}{20} - 0.3
\]

To exponentiate both sides of this equation, you will remove the logarithm by applying the base of the logarithm (assuming it's base 10 unless otherwise specified).

### Step-by-Step Exponentiation Process:

1. **Isolate the logarithmic part:**
   \[
   \log\left(\frac{r}{r_2}\right) = -\frac{6}{20} - 0.3
   \]
   First, simplify the right-hand side:
   \[
   -\frac{6}{20} = -0.3
   \]
   So, the equation becomes:
   \[
   \log\left(\frac{r}{r_2}\right) = -0.3 - 0.3 = -0.6
   \]

2. **Exponentiate both sides:**
   Since you are dealing with a logarithm of base 10, exponentiate both sides with base 10 to remove the logarithm:
   \[
   10^{\log\left(\frac{r}{r_2}\right)} = 10^{-0.6}
   \]
   This simplifies to:
   \[
   \frac{r}{r_2} = 10^{-0.6}
   \]

3. **Solve for \(r\):**
   Multiply both sides by \(r_2\) to isolate \(r\):
   \[
   r = r_2 \cdot 10^{-0.6}
   \]

Thus, the solution after exponentiating both sides is:

\[
r = r_2 \cdot 10^{-0.6}
\]

Do you want more details on any step or have any questions?

Here are five related questions to expand your understanding:

1. How do you handle logarithms with different bases (e.g., natural logs)?
2. What are the properties of logarithms that allow you to manipulate them algebraically?
3. How would you solve the equation if the logarithm was natural log \(\ln\) instead of \(\log\)?
4. What is the value of \(10^{-0.6}\) approximately?
5. How does exponentiation apply in real-world scenarios like sound intensity or pH levels?

**Tip:** Always check the base of the logarithm in your equation. If not explicitly given, it's usually base 10, but sometimes it's natural logarithm (base \(e\)).

How do you exponentiate both sides of this equation? log(r/r2) = -6/20 - 0.3

Learn how to exponentiate both sides of the logarithmic equation log(r/r2) = -6/20 - 0.3. This problem covers logarithmic and exponential algebraic concepts. Suitable for Grades 10-12 students.

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