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.