https://file.aiquickdraw.com/mathbot/file/1726593382106cwk2nqwa.jpg

The equation provided in the image is:

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

To solve this equation, let's apply the properties of logarithms.

### Step 1: Combine the logarithmic terms using the product rule for logarithms:
\[
\log(x) + \log(x + 3) = \log(x(x + 3)) = \log(x^2 + 3x)
\]

Now the equation becomes:
\[
\log(x^2 + 3x) = 1
\]

### Step 2: Convert the logarithmic equation to its exponential form:
\[
x^2 + 3x = 10^1 = 10
\]

### Step 3: Solve the quadratic equation:
\[
x^2 + 3x - 10 = 0
\]

We can solve this using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \(a = 1\), \(b = 3\), and \(c = -10\). Substituting these values into the quadratic formula:

\[
x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-10)}}{2(1)}
\]
\[
x = \frac{-3 \pm \sqrt{9 + 40}}{2}
\]
\[
x = \frac{-3 \pm \sqrt{49}}{2}
\]
\[
x = \frac{-3 \pm 7}{2}
\]

So, the two possible solutions are:

\[
x = \frac{-3 + 7}{2} = 2
\]
\[
x = \frac{-3 - 7}{2} = -5
\]

### Step 4: Verify the solutions:
- For \(x = 2\), both \(\log(2)\) and \(\log(2 + 3)\) are defined, so \(x = 2\) is a valid solution.
- For \(x = -5\), \(\log(-5)\) is undefined because the logarithm of a negative number does not exist in the set of real numbers. So, \(x = -5\) is not a valid solution.

### Final Solution:
The only valid solution is:
\[
x = 2
\]

Would you like more details or have any questions about this process?

Here are 5 related questions to expand your understanding:
1. How does the product rule of logarithms work, and why is it applicable here?
2. What are the steps to convert a logarithmic equation to an exponential one?
3. Could the quadratic formula always be used to solve any quadratic equation?
4. Why is the logarithm of a negative number undefined?
5. Can logarithmic equations have multiple solutions, and how do you check their validity?

**Tip:** Always check the domain of logarithmic functions to ensure that solutions are valid—logarithms are only defined for positive arguments.

Learn how to solve the logarithmic equation lg(x) + lg(x+3) = 1 using properties of logarithms and the quadratic formula. This step-by-step solution covers algebraic techniques and is suitable for high school students.

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