which of the following shows the true solution to the logarithmic equation below? log(x)+log(x + 5)=log(6x +…

Explanation:

Step1: Use log property

By the property $\log a+\log b=\log(ab)$, the left - hand side of the equation $\log(x)+\log(x + 5)=\log(x(x + 5))$. So the equation becomes $\log(x(x + 5))=\log(6x + 12)$.

Step2: Remove the log

Since the logarithm function $y = \log(u)$ is one - to - one, if $\log(u)=\log(v)$, then $u = v$. So $x(x + 5)=6x + 12$.

Step3: Expand and simplify

Expand the left - hand side: $x^{2}+5x=6x + 12$. Rearrange to get a quadratic equation: $x^{2}+5x-6x - 12=0$, which simplifies to $x^{2}-x - 12=0$.

Step4: Factor the quadratic

Factor $x^{2}-x - 12$ as $(x - 4)(x+3)=0$.

Step5: Solve for x

Set each factor equal to zero: $x - 4=0$ gives $x = 4$, and $x+3=0$ gives $x=-3$.

Step6: Check for domain

For $\log(x)$ to be well - defined, $x>0$. For $\log(x + 5)$ to be well - defined, $x+5>0$ (i.e., $x>-5$), and for $\log(6x + 12)$ to be well - defined, $6x+12>0$ (i.e., $x>-2$). When $x=-3$, $\log(x)$ is not defined. So we discard $x=-3$.

Answer:

B. $x = 4$