Sure, let's solve the equation step by step:
[tex]\[
\log_2(x + 6) = 3 - \log_2(x + 4)
\][/tex]
Step 1: Isolate the logarithmic terms on one side of the equation.
First, let's move [tex]\(\log_2(x + 4)\)[/tex] to the left side:
[tex]\[
\log_2(x + 6) + \log_2(x + 4) = 3
\][/tex]
Step 2: Combine the logarithmic terms using the properties of logarithms.
Recall the logarithmic property:
[tex]\[
\log_b(a) + \log_b(c) = \log_b(ac)
\][/tex]
Applying this property, we get:
[tex]\[
\log_2((x + 6)(x + 4)) = 3
\][/tex]
Step 3: Rewrite the equation in exponential form.
The definition of a logarithm tells us that if [tex]\(\log_b(a) = c\)[/tex], then [tex]\(a = b^c\)[/tex]. Applying this definition, we convert the logarithmic equation to an exponential equation:
[tex]\[
(x + 6)(x + 4) = 2^3
\][/tex]
Since [tex]\(2^3 = 8\)[/tex], we have:
[tex]\[
(x + 6)(x + 4) = 8
\][/tex]
Step 4: Expand and simplify the equation.
Expand the left side of the equation:
[tex]\[
x^2 + 4x + 6x + 24 = 8
\][/tex]
Combine like terms:
[tex]\[
x^2 + 10x + 24 = 8
\][/tex]
Subtract 8 from both sides to set the equation to zero:
[tex]\[
x^2 + 10x + 16 = 0
\][/tex]
Step 5: Solve the quadratic equation.
To factor [tex]\(x^2 + 10x + 16 = 0\)[/tex], we look for two numbers that multiply to 16 and add up to 10. These numbers are 2 and 8:
[tex]\[
(x + 2)(x + 8) = 0
\][/tex]
Set each factor to zero:
[tex]\[
x + 2 = 0 \quad \text{or} \quad x + 8 = 0
\][/tex]
Solving these equations gives us:
[tex]\[
x = -2 \quad \text{or} \quad x = -8
\][/tex]
Step 6: Check for extraneous solutions.
We must ensure that the solutions do not result in the logarithms of non-positive numbers in the original equation.
For [tex]\(x = -2\)[/tex]:
[tex]\[
x + 6 = -2 + 6 = 4 \quad (\text{valid}),
\][/tex]
[tex]\[
x + 4 = -2 + 4 = 2 \quad (\text{valid})
\][/tex]
For [tex]\(x = -8\)[/tex]:
[tex]\[
x + 6 = -8 + 6 = -2 \quad (\text{invalid, negative log}),
\][/tex]
[tex]\[
x + 4 = -8 + 4 = -4 \quad (\text{invalid, negative log})
\][/tex]
So, [tex]\(x = -8\)[/tex] results in taking the logarithm of a negative number, which is not defined.
Therefore, the only valid solution is:
[tex]\[
x = -2
\][/tex]
