Questions: Use the definition of a logarithm to solve the equation. A ln(9y) = ln(y^2 + 5y) y =

Solution Steps

To solve the equation \(\ln(9y) = \ln(y^2 + 5y)\), we can use the property of logarithms that states if \(\ln(a) = \ln(b)\), then \(a = b\). This allows us to set the arguments of the logarithms equal to each other: \(9y = y^2 + 5y\). We can then solve this resulting quadratic equation for \(y\).

We start with the equation given by the logarithmic equality: \[ \ln(9y) = \ln(y^2 + 5y) \] Using the property of logarithms that states if \(\ln(a) = \ln(b)\), then \(a = b\), we can set the arguments equal to each other: \[ 9y = y^2 + 5y \]

Rearranging the equation gives us: \[ y^2 + 5y - 9y = 0 \] which simplifies to: \[ y^2 - 4y = 0 \]

Factoring the quadratic equation, we have: \[ y(y - 4) = 0 \] Setting each factor equal to zero gives us the solutions: \[ y = 0 \quad \text{or} \quad y = 4 \]

Since \(y\) must be positive in the context of logarithms (as the logarithm of a non-positive number is undefined), we discard \(y = 0\). Thus, the only valid solution is: \[ y = 4 \]

Final Answer