In Exercises 79–84, solve for y.
83. ln(y-1) = x + ln(y)

Guida verificata passo dopo passo

Start with the given equation: \(\ln(y - 1) = x + \ln(y)\).

Use the property of logarithms that \(a + b = \ln(e^a) + \ln(e^b) = \ln(e^a \cdot e^b)\) to rewrite the right side: \(x + \ln(y) = \ln(e^x) + \ln(y) = \ln(y e^x)\).

Set the logarithmic expressions equal: \(\ln(y - 1) = \ln(y e^x)\).

Since the natural logarithm function is one-to-one, equate the arguments: \(y - 1 = y e^x\).

Solve the resulting equation for \(y\) by isolating \(y\) on one side: \(y - y e^x = 1\), then factor \(y\) and solve for it.

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Properties of Logarithms

Understanding the properties of logarithms, such as the product, quotient, and power rules, is essential. These properties allow you to combine or separate logarithmic expressions, which is crucial for simplifying and solving equations involving logarithms.

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Solving Logarithmic Equations

Solving logarithmic equations often involves isolating the logarithmic terms and then exponentiating both sides to eliminate the logarithm. This process transforms the equation into an algebraic form that can be solved for the variable.

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Solving Logarithmic Equations

Domain Restrictions in Logarithmic Functions

Logarithmic functions are only defined for positive arguments. When solving equations like ln(y-1) = x + ln(y), it is important to consider domain restrictions such as y > 1 and y > 0 to ensure the solution is valid.

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