Question

Solve the equation $\log \left(3 x^{2}-24\right)=\log \left(2 x^{2}-2 x\right)$. Explain your work in 3-5 sentences.

Answer 1

We start by equating the arguments of the logarithms since \(\log(a) = \log(b)\) implies \(a = b\) (as long as \(a > 0\) and \(b > 0\)). Thus, we set up the equation:

\[
3x^2 - 24 = 2x^2 - 2x.
\]

Next, we rearrange it to one side:

\[
3x^2 - 2x^2 + 2x - 24 = 0 \Rightarrow x^2 + 2x - 24 = 0.
\]

Now we can factor the quadratic equation:

\[
(x + 6)(x - 4) = 0.
\]

This gives us the solutions \(x = -6\) and \(x = 4\). However, we must check that these solutions keep the arguments of the logarithms positive.

For \(x = -6\):

\[
3(-6)^2 - 24 = 108 - 24 = 84 > 0,
\]
\[
2(-6)^2 - 2(-6) = 72 + 12 = 84 > 0.
\]

Both arguments are valid.

For \(x = 4\):

\[
3(4)^2 - 24 = 48 - 24 = 24 > 0,
\]
\[
2(4)^2 - 2(4) = 32 - 8 = 24 > 0.
\]

Both arguments are also valid. Hence, the solutions to the equation are \(x = -6\) and \(x = 4\).