which of the following shows the true solution to the logarithmic equation solved below?\nlog₂(x)+log₂(x +…

which of the following shows the true solution to the logarithmic equation solved below?\nlog₂(x)+log₂(x + 7)=3\nlog₂x(x + 7)=3\nx(x + 7)=2³\nx²+7x - 8=0\n(x + 8)(x - 1)=0\nx=-8,1\no x=-8\no x=1\no x=1 and x=-8\no x=1 and x=8

which of the following shows the true solution to the logarithmic equation solved below?\nlog₂(x)+log₂(x + 7)=3\nlog₂x(x + 7)=3\nx(x + 7)=2³\nx²+7x - 8=0\n(x + 8)(x - 1)=0\nx=-8,1\no x=-8\no x=1\no x=1 and x=-8\no x=1 and x=8

Answer

Explanation:

Step1: Recall log - product rule

Using the rule $\log_a(M)+\log_a(N)=\log_a(MN)$, we get $\log_2[x(x + 7)]=3$ from $\log_2(x)+\log_2(x + 7)=3$.

Step2: Convert to exponential form

By the definition of logarithms, if $\log_a(b)=c$, then $b=a^c$. So $x(x + 7)=2^3=8$, which simplifies to $x^{2}+7x - 8=0$.

Step3: Factor the quadratic equation

Factoring $x^{2}+7x - 8=0$ gives $(x + 8)(x - 1)=0$, so $x=-8$ or $x = 1$.

Step4: Check for valid solutions

For the original logarithmic equation $\log_2(x)+\log_2(x + 7)=3$, the argument of a logarithm must be positive. When $x=-8$, $\log_2(-8)$ and $\log_2(-8 + 7)=\log_2(-1)$ are undefined. When $x = 1$, $\log_2(1)=0$ and $\log_2(1+7)=\log_2(8)=3$, and the original equation holds.

Answer:

$x = 1$