https://file.aiquickdraw.com/mathbot/file/17295756900307g4vtyje.jpg

The equation provided in the image is:

\[
\log(x - 3) - \log(x - 2) = 2
\]

We can solve this step by step.

### Step 1: Use the logarithmic property
The difference of two logarithms can be written as the logarithm of a quotient:
\[
\log\left(\frac{x - 3}{x - 2}\right) = 2
\]

### Step 2: Exponentiate both sides
To eliminate the logarithm, exponentiate both sides with base 10:
\[
\frac{x - 3}{x - 2} = 10^2 = 100
\]

### Step 3: Solve for \( x \)
Now, solve the equation:
\[
\frac{x - 3}{x - 2} = 100
\]

Multiply both sides by \( x - 2 \) to get rid of the fraction:
\[
x - 3 = 100(x - 2)
\]

Expand the right-hand side:
\[
x - 3 = 100x - 200
\]

Move all terms involving \( x \) to one side and constants to the other:
\[
x - 100x = -200 + 3
\]
\[
-99x = -197
\]

Solve for \( x \):
\[
x = \frac{197}{99} \approx 1.99
\]

### Step 4: Check the solution
For a logarithmic equation, the arguments inside the logarithms must be positive. So we check:

- \( x - 3 > 0 \) gives \( x > 3 \)
- \( x - 2 > 0 \) gives \( x > 2 \)

Since \( x = \frac{197}{99} \approx 1.99 \), it does not satisfy the condition \( x > 3 \) or \( x > 2 \). Therefore, **there are no valid solutions** for this equation.

Would you like a further breakdown or clarification?

### Related Questions:
1. What is the general rule for solving logarithmic equations involving subtraction?
2. How can you determine the domain of a logarithmic function?
3. How do you handle cases where a logarithmic equation has no solution?
4. What are the implications of extraneous solutions in logarithmic equations?
5. How does the base of the logarithm affect the steps in solving the equation?

**Tip:** Always check the domain restrictions when working with logarithmic functions to avoid extraneous solutions.

Step-by-step solution of the logarithmic equation log(x - 3) - log(x - 2) = 2. In this problem, key concepts such as properties of logarithms and solving equations using logarithms are covered. Suitable for Grades 10-12.

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