Question

which of the following is true regarding the solutions to the logarithmic equation below?
$2\log_{3}(x) = 4$
$log_{3}(x^{2}) = 4$
\quad\quad$x^{2} = 3^{4}$
\quad\quad$x^{2} = 81$
\quad\quad$x = 9, -9$
\bigcirc $x = 9$ and $x = -9$ are true solutions
\bigcirc $x = 9$ and $x = -9$ are extraneous solutions
\bigcirc $x = 9$ is an extraneous solution and $x = -9$ is a true solution
\bigcirc $x = 9$ is a true solution and $x = -9$ is an extraneous solution

Explanation:

Step1: Recall the domain of logarithmic functions

The argument of a logarithm, like \(\log_b(a)\), must satisfy \(a > 0\) (since you can't take the logarithm of a non - positive number). For the original equation \(2\log_3(x)=4\), the argument of the logarithm is \(x\), so \(x>0\).

Step2: Check \(x = 9\)

Substitute \(x = 9\) into the original equation \(2\log_3(x)\). We know that \(\log_3(9)=\log_3(3^2) = 2\) (by the property \(\log_a(a^b)=b\)). Then \(2\log_3(9)=2\times2 = 4\), which matches the right - hand side of the original equation. So \(x = 9\) is a valid solution.

Step3: Check \(x=-9\)

Substitute \(x=-9\) into the original equation \(2\log_3(x)\). But the argument of the logarithm \(\log_3(-9)\) is not defined (because the argument of a logarithm must be positive). So even though \(x = - 9\) satisfies the equation \(x^{2}=81\), it does not satisfy the original logarithmic equation because it makes the argument of the logarithm non - positive. So \(x=-9\) is an extraneous solution.

Answer:

\(x = 9\) is a true solution and \(x=-9\) is an extraneous solution (the option: \(x = 9\) is a true solution and \(x=-9\) is an extraneous solution)