Question

solve the logarithmic equation. be sure to reject any value of x that is not in the domain of the original logarithmic expressions.\\(\log_{4}x + \log_{4}(3x - 1) = 1\\)\
select the correct choice below and, if necessary, fill in the answer box to complete your choice.\
a. the solution set is \\(\boxed{}\\)\
(type an exact answer in simplified form. use a comma to separate answers as needed.)\
b. there is no solution.

Explanation:

Step1: Apply log product rule

Using the property \(\log_b M + \log_b N=\log_b(MN)\), we combine the left - hand side of the equation \(\log_4 x+\log_4(3x - 1)=1\). So we get \(\log_4[x(3x - 1)] = 1\).

Step2: Convert to exponential form

Recall that if \(\log_b a=c\), then \(b^c=a\). For our equation \(\log_4[x(3x - 1)] = 1\), where \(b = 4\), \(c = 1\) and \(a=x(3x - 1)\), we can rewrite it in exponential form as \(4^1=x(3x - 1)\).

Step3: Solve the quadratic equation

Simplify the right - hand side: \(4 = 3x^2-x\). Rearrange it to the standard quadratic form \(3x^2-x - 4=0\).
We factor the quadratic equation: \(3x^2-x - 4=(3x - 4)(x + 1)=0\).
Set each factor equal to zero:

• For \(3x-4 = 0\), we have \(3x=4\), so \(x=\frac{4}{3}\).
• For \(x + 1=0\), we have \(x=-1\).

Step4: Check the domain

The domain of the logarithmic function \(\log_4 x\) requires \(x>0\), and for \(\log_4(3x - 1)\), we need \(3x-1>0\), which implies \(x>\frac{1}{3}\).

• When \(x=-1\), \(\log_4(-1)\) and \(\log_4(3\times(-1)-1)=\log_4(-4)\) are not defined (since the argument of a logarithm must be positive).
• When \(x = \frac{4}{3}\), \(\log_4\frac{4}{3}\) is defined (because \(\frac{4}{3}>0\)) and \(\log_4(3\times\frac{4}{3}-1)=\log_4(4 - 1)=\log_4 3\) is also defined (because \(3>0\)).

Answer:

\(\frac{4}{3}\)