Question

question 7 (1 point) solve the equation \\(\log(x + 6) - \log(2x + 1) = \log 4\\). \\(\bigcirc\\) a) \\(x = \frac{9}{2}\\) \\(\bigcirc\\) b) \\(x = \frac{2}{7}\\) \\(\bigcirc\\) c) \\(x = 1\\) \\(\bigcirc\\) d) \\(x = \frac{5}{7}\\)

Explanation:

Step1: Apply log quotient rule

Using the property \(\log a - \log b=\log\frac{a}{b}\), rewrite the left - hand side:
\(\log\frac{x + 6}{2x+1}=\log4\)

Step2: Set arguments equal (since log is one - to - one)

If \(\log m=\log n\), then \(m = n\) (for the same base, and \(m>0,n>0\)). So we have:
\(\frac{x + 6}{2x + 1}=4\)

Step3: Solve the resulting equation

Multiply both sides by \(2x + 1\) (assuming \(2x+1
eq0\), i.e., \(x
eq-\frac{1}{2}\)):
\(x + 6=4(2x + 1)\)
Expand the right - hand side:
\(x + 6 = 8x+4\)
Subtract \(x\) from both sides:
\(6=7x + 4\)
Subtract 4 from both sides:
\(2 = 7x\)
Divide both sides by 7:
\(x=\frac{2}{7}\)
We also need to check the domain:
For \(\log(x + 6)\), \(x+6>0\Rightarrow x>-6\)
For \(\log(2x + 1)\), \(2x + 1>0\Rightarrow x>-\frac{1}{2}\)
For \(x = \frac{2}{7}\), \(\frac{2}{7}>- \frac{1}{2}\) and \(\frac{2}{7}+6=\frac{2 + 42}{7}=\frac{44}{7}>0\), \(2\times\frac{2}{7}+1=\frac{4 + 7}{7}=\frac{11}{7}>0\). So \(x=\frac{2}{7}\) is a valid solution.

Answer:

b) \(x=\frac{2}{7}\)