https://file.aiquickdraw.com/mathbot/file/1725685962465s3hghf4u.jpg

The image contains the following logarithmic equation:

\[
\log_3(x + 25) - \log_3(x - 1) = 3
\]

Let's solve this step by step.

### Step 1: Apply the properties of logarithms
Using the property \(\log_b A - \log_b B = \log_b \left(\frac{A}{B}\right)\), we can rewrite the equation as:

\[
\log_3 \left( \frac{x + 25}{x - 1} \right) = 3
\]

### Step 2: Eliminate the logarithm
To eliminate the logarithm, rewrite the equation in exponential form. Since \(\log_3 y = z\) implies \(y = 3^z\), we get:

\[
\frac{x + 25}{x - 1} = 3^3
\]

\[
\frac{x + 25}{x - 1} = 27
\]

### Step 3: Solve the equation
Now, solve for \(x\) by multiplying both sides by \(x - 1\):

\[
x + 25 = 27(x - 1)
\]

Expand the right side:

\[
x + 25 = 27x - 27
\]

### Step 4: Simplify the equation
Now, rearrange the terms to isolate \(x\):

\[
25 + 27 = 27x - x
\]

\[
52 = 26x
\]

\[
x = \frac{52}{26} = 2
\]

### Step 5: Check for extraneous solutions
We must check that \(x = 2\) doesn't make any logarithms undefined. Substituting \(x = 2\):

- \(x + 25 = 27\)
- \(x - 1 = 1\)

Both values are positive, so the solution \(x = 2\) is valid.

Thus, the solution is \(x = 2\).

Do you have any questions or would you like more details on any step? Here are some related questions for further exploration:

1. What happens when the base of the logarithm is changed?
2. How do you solve logarithmic equations with different bases?
3. Can you generalize this approach to equations with more logarithmic terms?
4. How do you solve logarithmic inequalities?
5. What are the conditions for the existence of logarithmic solutions?

**Tip:** Always check for extraneous solutions when solving logarithmic equations by verifying that all logarithmic terms are defined (i.e., their arguments are positive).

This solution explains how to solve the logarithmic equation log_3(x + 25) - log_3(x - 1) = 3 step by step using properties of logarithms and exponential equations. The problem is suitable for students in Grades 10-12.

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