Question

which of the following is true regarding the solutions to the logarithmic equation below?
$2\log_{6}(x) = 2$
$log_{6}(x^{2}) = 2$
\quad\quad$x^{2} = 6^{2}$
\quad\quad$x^{2} = 36$
\quad\quad$x = 6, -6$
\bigcirc $x = 6$ and $x = -6$ are true solutions
\bigcirc $x = 6$ and $x = -6$ are extraneous solutions
\bigcirc $x = 6$ is a true solution and $x = -6$ is an extraneous solution
\bigcirc $x = 6$ is an extraneous solution and $x = -6$ is a true solution

Explanation:

Step1: Recall domain of logarithm

The argument of a logarithm, $\log_b(a)$, must satisfy $a > 0$. For the original equation $2\log_6(x) = 2$, the argument is $x$, so $x > 0$.

Step2: Test $x = 6$

Substitute $x = 6$ into the original equation: $2\log_6(6)=2\log_6(6)=2\times1 = 2$, which holds true.

Step3: Test $x = -6$

Substitute $x = -6$ into the original equation: The argument of $\log_6(-6)$ is negative, which is undefined in the real - valued logarithm function. So $x=-6$ is not a valid solution (it is an extraneous solution).

Answer:

C. $x = 6$ is a true solution and $x=-6$ is an extraneous solution (assuming the options are labeled as A, B, C, D with C being the correct one as per the analysis)