https://file.aiquickdraw.com/mathbot/file/1728725744518i9o9hb7k.jpg

The equation from the image is:

\[
\log_5(2x - 3) + 2\log_5 2 = 1 + \log_5(3 - 2x)
\]

### Steps to solve the equation:

1. **Apply the logarithmic property**: 
   The second term, \(2\log_5 2\), can be rewritten using the power rule of logarithms:
   \[
   2\log_5 2 = \log_5(2^2) = \log_5 4
   \]
   So, the equation becomes:
   \[
   \log_5(2x - 3) + \log_5 4 = 1 + \log_5(3 - 2x)
   \]

2. **Combine the logarithms**: 
   On the left-hand side, we use the addition property of logarithms:
   \[
   \log_5(2x - 3) + \log_5 4 = \log_5[(2x - 3) \cdot 4] = \log_5(4(2x - 3))
   \]
   This simplifies the equation to:
   \[
   \log_5(4(2x - 3)) = 1 + \log_5(3 - 2x)
   \]

3. **Isolate the logarithmic terms**:
   Subtract \(\log_5(3 - 2x)\) from both sides:
   \[
   \log_5\left(\frac{4(2x - 3)}{3 - 2x}\right) = 1
   \]

4. **Convert the logarithmic equation to exponential form**:
   Recall that \( \log_b a = c \) means \( a = b^c \). In this case:
   \[
   \frac{4(2x - 3)}{3 - 2x} = 5^1 = 5
   \]

5. **Solve for \(x\)**:
   Now, solve the equation:
   \[
   \frac{4(2x - 3)}{3 - 2x} = 5
   \]
   Multiply both sides by \((3 - 2x)\) to eliminate the denominator:
   \[
   4(2x - 3) = 5(3 - 2x)
   \]
   Expand both sides:
   \[
   8x - 12 = 15 - 10x
   \]
   Bring all terms involving \(x\) to one side:
   \[
   8x + 10x = 15 + 12
   \]
   \[
   18x = 27
   \]
   Solve for \(x\):
   \[
   x = \frac{27}{18} = \frac{3}{2} = 1.5
   \]

### Check for extraneous solutions:
We need to verify that \(x = 1.5\) doesn't make the arguments of the logarithms negative.

- \(2x - 3 = 2(1.5) - 3 = 0\) (logarithms are undefined for 0 or negative values)
- \(3 - 2x = 3 - 2(1.5) = 0\)

Since the logarithms involve 0 at \(x = 1.5\), this is an extraneous solution.

### Conclusion:
There is **no solution** to this equation because \(x = 1.5\) leads to an invalid logarithmic expression.

---

Would you like more details on any step, or do you have any further questions?

Here are 5 related questions for further exploration:
1. How do you solve equations involving logarithms of different bases?
2. What are common logarithmic properties and how do they simplify equations?
3. What are extraneous solutions, and why do they occur in logarithmic equations?
4. Can this equation be solved using a different base for logarithms, like base 10?
5. How do logarithmic and exponential functions relate to each other?

**Tip**: Always check the domain of logarithmic expressions to avoid extraneous solutions. Logarithms are only defined for positive arguments.

log_5(2x - 3) + 2log_5 2 = 1 + log_5(3 - 2x)

This problem involves solving a logarithmic equation with base 5: log_5(2x - 3) + 2log_5 2 = 1 + log_5(3 - 2x). The solution covers properties of logarithms and addresses extraneous solutions. Suitable for students in Grades 10-12.

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