To solve the equation
[tex]\[
\log_4(x + 3) + \log_4(x) = 1,
\][/tex]
we will proceed step-by-step.
### Step 1: Use the properties of logarithms
We know the following property of logarithms for the same base:
[tex]\[
\log_b(a) + \log_b(c) = \log_b(a \cdot c).
\][/tex]
Applying this property to our equation, we get:
[tex]\[
\log_4(x + 3) + \log_4(x) = \log_4((x + 3) \cdot x).
\][/tex]
So, the equation becomes:
[tex]\[
\log_4(x(x + 3)) = 1.
\][/tex]
### Step 2: Simplify inside the logarithm
Next, simplify the expression inside the logarithm:
[tex]\[
x(x + 3) = x^2 + 3x.
\][/tex]
So, our equation now reads:
[tex]\[
\log_4(x^2 + 3x) = 1.
\][/tex]
### Step 3: Apply the definition of logarithms
By definition, [tex]\(\log_b(a) = c\)[/tex] implies [tex]\(b^c = a\)[/tex]. Therefore:
[tex]\[
4^1 = x^2 + 3x.
\][/tex]
Simplifying the right-hand side:
[tex]\[
4 = x^2 + 3x.
\][/tex]
### Step 4: Form a quadratic equation
This translates to a quadratic equation:
[tex]\[
x^2 + 3x - 4 = 0.
\][/tex]
### Step 5: Solve the quadratic equation
To solve the quadratic equation [tex]\(x^2 + 3x - 4 = 0\)[/tex], we either factorize it or use the quadratic formula. In this case, it can be factorized:
[tex]\[
(x + 4)(x - 1) = 0.
\][/tex]
This gives us the solutions:
[tex]\[
x + 4 = 0 \quad \text{or} \quad x - 1 = 0,
\][/tex]
which means:
[tex]\[
x = -4 \quad \text{or} \quad x = 1.
\][/tex]
### Step 6: Verify solutions
Check whether both solutions are valid in the context of the original logarithmic equation. Recall that logarithms are only defined for positive arguments.
1. For [tex]\(x = 1\)[/tex]:
[tex]\[
\log_4(1 + 3) + \log_4(1) = \log_4(4) + \log_4(1) = 1 + 0 = 1.
\][/tex]
This is indeed true.
2. For [tex]\(x = -4\)[/tex]:
[tex]\[
\log_4(-4 + 3) + \log_4(-4) = \log_4(-1) + \log_4(-4),
\][/tex]
which is undefined because we cannot take the logarithm of a negative number in the real number system.
Thus, [tex]\(x = -4\)[/tex] is not a valid solution.
### Conclusion
The only valid solution to the equation is:
[tex]\[
\boxed{x = 1}.
\][/tex]
