Find Inverse of Logarithmic Functions

• Domain of \(f\): \(x - 2 > 0 \Rightarrow x > 2\), so \(\text{Domain}(f) = (2, +\infty)\) and \(\text{Range}(f) = (-\infty, +\infty)\).
• Write the function as an equation: \[ y = \ln(x - 2) \]
• Convert to exponential form: \[ x - 2 = e^y \]
• Solve for \(x\): \[ x = 2 + e^y \]
• Swap \(x\) and \(y\) to get the inverse function: \[ f^{-1}(x) = 2 + e^x \]

Conclusion: \(\text{Domain}(f^{-1}) = (-\infty, +\infty)\), \(\text{Range}(f^{-1}) = (2, +\infty)\).

• Domain of \(f\): \(4x - 6 > 0 \Rightarrow x > \frac{3}{2}\), so \(\text{Domain}(f) = \left(\frac{3}{2}, +\infty\right)\), \(\text{Range}(f) = (-\infty, +\infty)\).
• Equation form and isolation: \[ y = 3\ln(4x - 6) - 2 \implies \ln(4x - 6) = \frac{y + 2}{3} \]
• Exponential form: \[ 4x - 6 = e^{(y + 2)/3} \]
• Solve for \(x\): \[ 4x = e^{(y + 2)/3} + 6 \implies x = \frac{1}{4} e^{(y + 2)/3} + \frac{3}{2} \]
• Swap variables for the inverse function: \[ f^{-1}(x) = \frac{1}{4} e^{(x + 2)/3} + \frac{3}{2} \]

Conclusion: \(\text{Domain}(f^{-1}) = (-\infty, +\infty)\), \(\text{Range}(f^{-1}) = \left(\frac{3}{2}, +\infty\right)\).

• Domain and Range of \(f\): \(\text{Domain}(f) = (-\infty, -2)\), \(\text{Range}(f) = (-\infty, +\infty)\).
• Equation form: \[ y = -\ln(x^2 - 4) - 5 \implies \ln(x^2 - 4) = -y - 5 \]
• Exponential form: \[ x^2 - 4 = e^{-y - 5} \implies x^2 = e^{-y - 5} + 4 \implies x = \pm \sqrt{e^{-y - 5} + 4} \]
• Select root using domain restriction (\(x < -2\)): \[ x = - \sqrt{e^{-y - 5} + 4} \]
• Inverse function: \[ f^{-1}(x) = - \sqrt{e^{-x - 5} + 4} \]

Conclusion: \(\text{Domain}(f^{-1}) = (-\infty, +\infty)\), \(\text{Range}(f^{-1}) = (-\infty, -2)\).