https://file.aiquickdraw.com/mathbot/file/1725695072839vk7iscsg.jpg

The equation extracted from the image seems to be:

\[
\log x + \log (x - 3) = 1
\]

Let's solve this equation.

### Step-by-Step Solution:

1. Use the logarithmic property:
   \[
   \log a + \log b = \log(ab)
   \]
   Applying this, the equation becomes:
   \[
   \log [x(x - 3)] = 1
   \]

2. Convert from logarithmic to exponential form:
   \[
   x(x - 3) = 10^1
   \]
   Simplifying:
   \[
   x(x - 3) = 10
   \]
   \[
   x^2 - 3x = 10
   \]

3. Rearrange the equation to form a quadratic equation:
   \[
   x^2 - 3x - 10 = 0
   \]

4. Solve the quadratic equation using the quadratic formula:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   For \( a = 1 \), \( b = -3 \), and \( c = -10 \), we substitute these values into the formula.

I'll compute the solutions now.The solutions to the quadratic equation \(x^2 - 3x - 10 = 0\) are:

\[
x_1 = 5 \quad \text{and} \quad x_2 = -2
\]

However, since logarithms are only defined for positive numbers, we must discard the negative solution \(x = -2\). Therefore, the valid solution is:

\[
x = 5
\]

Let me know if you'd like more details on any of the steps! Here are a few related questions:

1. How do you verify if the solution \(x = 5\) satisfies the original equation?
2. What happens if both solutions of the quadratic were positive?
3. Can you apply similar logarithmic properties to solve more complex equations?
4. How does the base of the logarithm affect the solution process?
5. What if the equation had an inequality, such as \(\log x + \log(x-3) > 1\)?

**Tip:** Always check the domain of the logarithmic function when solving such equations to avoid extraneous solutions.

Learn how to solve the logarithmic equation log x + log(x - 3) = 1 using logarithmic properties and quadratic formulas. This detailed solution is suitable for high school students in Grades 9-12.

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