Question

which of the following correctly solves the equation \\(\log_{3}(2x - 1) = \log_{3}(3x)\\)? (1 point) \\(\bigcirc\\) \\(x = -1\\) \\(\bigcirc\\) \\(x = \frac{1}{5}\\) \\(\bigcirc\\) \\(x = 1\\) \\(\bigcirc\\) no solution

Explanation:

Step1: Use logarithmic property

If \(\log_{a}M = \log_{a}N\), then \(M = N\) (for \(a>0,a
eq1,M>0,N>0\)). So we set \(2x - 1=3x\).

Step2: Solve the equation

Subtract \(2x\) from both sides: \(2x - 1-2x=3x - 2x\), which simplifies to \(- 1=x\), so \(x = - 1\).

Step3: Check domain

For \(\log_{3}(2x - 1)\), we need \(2x - 1>0\), substituting \(x=-1\) gives \(2(-1)-1=-3<0\). For \(\log_{3}(3x)\), substituting \(x = - 1\) gives \(3(-1)=-3<0\). Both arguments are negative, so \(x=-1\) is not in the domain.

Answer:

no solution