Question

solve the equation \\(\frac{ln(x + 7)}{ln(2x - 2)} = 1\\). if no solutions exists, enter none.

Explanation:

Step1: Cross - multiply to simplify the equation

Given the equation \(\frac{\ln(x + 7)}{\ln(2x-2)}=1\). Since the denominator \(\ln(2x - 2)
eq0\) (because division by zero is undefined), we can multiply both sides of the equation by \(\ln(2x - 2)\) to get \(\ln(x + 7)=\ln(2x - 2)\).

Step2: Use the property of logarithms

If \(\ln a=\ln b\), then \(a = b\) (for the natural logarithm, the function \(y = \ln x\) is one - to - one, which means that if \(\ln m=\ln n\), then \(m=n\) as long as \(m>0\) and \(n>0\)). So we have the equation \(x + 7=2x - 2\).

Step3: Solve the linear equation

Subtract \(x\) from both sides of the equation \(x + 7=2x - 2\):
\(x+7 - x=2x - 2 - x\)
\(7=x - 2\)
Then add 2 to both sides:
\(x=7 + 2=9\)

Step4: Check the domain of the original logarithmic functions

For the function \(y = \ln(x + 7)\), we need \(x+7>0\), i.e., \(x>- 7\). For the function \(y=\ln(2x - 2)\), we need \(2x-2>0\), which simplifies to \(2x>2\) or \(x > 1\). Also, we need to ensure that \(\ln(2x - 2)
eq0\). If \(\ln(2x - 2)=0\), then \(2x-2 = 1\) (since \(\ln1 = 0\)), \(2x=3\), \(x=\frac{3}{2}\). When \(x = 9\), \(x+7=16>0\), \(2x - 2=18 - 2 = 16>0\), and \(\ln(2x - 2)=\ln(16)
eq0\).

Answer:

\(x = 9\)