Question

solve the logarithmic equation. be sure to reject any value of x that is not in the domain of the original logarithmic expressions. give an exact answer. \\(\log_{12}x + \log_{12}(11x - 1) = 1\\) solve the equation. 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 integers or fractions for any numbers in the expression.) b. there are infinitely many solutions. c. 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 \(\log_{12}x+\log_{12}(11x - 1)=\log_{12}[x(11x - 1)]\). So the equation becomes \(\log_{12}[x(11x - 1)] = 1\).

Step2: Convert to exponential form

Recall that if \(\log_b y=k\), then \(y = b^k\). For our equation, \(b = 12\), \(k = 1\) and \(y=x(11x - 1)\). So we have \(x(11x - 1)=12^1=12\).

Step3: Expand and form quadratic equation

Expand the left - hand side: \(11x^{2}-x = 12\). Rearrange it to the standard quadratic form \(ax^{2}+bx + c = 0\): \(11x^{2}-x - 12=0\).

Step4: Solve the quadratic equation

We can factor the quadratic equation. We need two numbers that multiply to \(11\times(- 12)=-132\) and add up to \(-1\). The numbers are \(-12\) and \(11\). So we rewrite the middle term:
\(11x^{2}+11x-12x - 12 = 0\)
Factor by grouping:
\(11x(x + 1)-12(x + 1)=0\)
\((11x - 12)(x + 1)=0\)

Set each factor equal to zero:

• For \(11x-12 = 0\), we get \(x=\frac{12}{11}\).
• For \(x + 1=0\), we get \(x=-1\).

Step5: Check the domain

The domain of a logarithmic function \(\log_b u\) requires that \(u>0\).

• For \(x=-1\):
• When \(x=-1\), in \(\log_{12}x\), \(x=-1<0\), so \(x = - 1\) is not in the domain.
• For \(x=\frac{12}{11}\):
• When \(x = \frac{12}{11}\), \(x=\frac{12}{11}>0\) and \(11x-1=11\times\frac{12}{11}-1=12 - 1 = 11>0\). So \(x=\frac{12}{11}\) is in the domain.

Answer:

\(\frac{12}{11}\)