Question

solve the equation \\(dfrac{ln(x + 2)}{ln(2x - 3)} = 1\\). if no solutions exists, enter none.

Explanation:

Step1: Analyze the domain

For the logarithms to be defined, we need \(x + 2>0\) (so \(x>- 2\)) and \(2x-3>0\) (so \(x>\frac{3}{2}\)). Also, since the denominator \(\ln(2x - 3)
eq0\), we have \(2x-3
eq1\) (because \(\ln(1) = 0\)), so \(2x
eq4\) or \(x
eq2\). The domain is \(x>\frac{3}{2}\) and \(x
eq2\).

Step2: Solve the equation

Given \(\frac{\ln(x + 2)}{\ln(2x-3)}=1\), we can multiply both sides by \(\ln(2x - 3)\) (since \(\ln(2x - 3)
eq0\) from the domain) to get \(\ln(x + 2)=\ln(2x - 3)\).

Since the natural logarithm function \(y = \ln(u)\) is one - to - one (if \(\ln(a)=\ln(b)\), then \(a = b\) for \(a>0\) and \(b>0\)), we have \(x + 2=2x-3\).

Solving \(x + 2=2x-3\):
Subtract \(x\) from both sides: \(2=x - 3\).
Add 3 to both sides: \(x=5\).

Step3: Check the solution in the domain

We check if \(x = 5\) is in the domain. For \(x = 5\), \(x+2=5 + 2=7>0\), \(2x-3=2\times5-3 = 10 - 3=7>0\), and \(\ln(2x - 3)=\ln(7)
eq0\). So \(x = 5\) is in the domain.

Answer:

\(x = 5\)