Which of the following shows the true solution to the logarithmic equation solved below?

[tex]\[
\begin{aligned}
\log_2(x) + \log_2(x+7) &= 3 \\
\log_2(x(x+7)) &= 3 \\
x(x+7) &= 2^3 \\
x^2 + 7x - 8 &= 0 \\
(x + 8)(x - 1) &= 0 \\
x &= -8, 1
\end{aligned}
\][/tex]

A. \( x = -8 \)

B. \( x = 1 \)

C. \( x = 1 \) and \( x = -8 \)

D. [tex]\( x = 1 \)[/tex] and [tex]\( x = 8 \)[/tex]



Answer :

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]