Question

7 solve the following equation: \\(log_{4} (3x - 1) = log_{4} (2x + 3)\\) \\(\boldsymbol{\text{a}}\\) 4 \\(\boldsymbol{\text{b}}\\) \\(\frac{4}{5}\\) \\(\boldsymbol{\text{c}}\\) no solution \\(\boldsymbol{\text{d}}\\) 2

Explanation:

Step1: Use Logarithm Property

For logarithmic equations with the same base, if $\log_b(M) = \log_b(N)$, then $M = N$ (where $b>0, b
eq1, M>0, N>0$). Here, base $b = 4$, so we set $3x - 1 = 2x + 3$.

Step2: Solve for x

Subtract $2x$ from both sides: $3x - 2x - 1 = 2x - 2x + 3$ → $x - 1 = 3$. Then add 1 to both sides: $x - 1 + 1 = 3 + 1$ → $x = 4$.

Step3: Check Domain

For $\log_4(3x - 1)$, when $x = 4$, $3(4) - 1 = 11 > 0$. For $\log_4(2x + 3)$, when $x = 4$, $2(4) + 3 = 11 > 0$. Both are in the domain.

Answer:

A. 4