To solve the given logarithmic equation step-by-step, we start with:
[tex]\[
\log_2(x) + \log_2(x + 7) = 3
\][/tex]
Using the properties of logarithms, specifically the product rule \( \log_b(a) + \log_b(c) = \log_b(ac) \), we can combine the logarithms:
[tex]\[
\log_2(x(x + 7)) = 3
\][/tex]
Next, we rewrite the equation in its exponential form. Recall that \( \log_b(a) = c \) implies \( a = b^c \):
[tex]\[
x(x + 7) = 2^3
\][/tex]
Calculate \( 2^3 \):
[tex]\[
x(x + 7) = 8
\][/tex]
Now, we rearrange this into a standard quadratic equation:
[tex]\[
x^2 + 7x - 8 = 0
\][/tex]
Next, solve this quadratic equation by factoring. We find two numbers that multiply to \(-8\) and add to \(7\):
[tex]\[
(x + 8)(x - 1) = 0
\][/tex]
Setting each factor equal to zero gives us the potential solutions:
[tex]\[
x + 8 = 0 \quad \Rightarrow \quad x = -8
\][/tex]
[tex]\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\][/tex]
We must now check the domain of the original logarithmic functions. Logarithms are only defined for positive arguments:
[tex]\[
\log_2(x) \quad \text{is defined for} \quad x > 0 \\
\log_2(x + 7) \quad \text{is defined for} \quad x + 7 > 0 \quad \Rightarrow \quad x > -7
\][/tex]
From these conditions, \( x = -8 \) is not valid because the argument of the logarithm would be negative. Thus, the only valid solution is:
[tex]\[
x = 1
\][/tex]
Therefore, the true solution to the logarithmic equation is:
[tex]\[
x = 1
\][/tex]
So, the following shows the true solution:
[tex]\[
x = 1
\][/tex]
