Questions: Use the one-to-one property of logarithms to solve. List all working and extraneous solutions. [ ln (-4 x)=ln left(x^2-3 xright) ] (Use commas to separate multiple answers, or type "NONE" if appropriate.) Solution(s): x= Extraneous: x ≠

Use the one-to-one property of logarithms to solve. List all working and extraneous solutions.
[
ln (-4 x)=ln left(x^2-3 xright)
]
(Use commas to separate multiple answers, or type "NONE" if appropriate.)
Solution(s): x=
Extraneous: x ≠

Transcript text: Use the one-to-one property of logarithms to solve. List all working and extraneous solutions. \[ \ln (-4 x)=\ln \left(x^{2}-3 x\right) \] (Use commas to separate multiple answers, or type "NONE" if appropriate.) Solution(s): $x=$ $\square$ Extraneous: $x \neq$ $\square$

Solution

Solution Steps

To solve the equation $\ln (-4x) = \ln (x^2 - 3x)$ using the one-to-one property of logarithms, we can set the arguments of the logarithms equal to each other and solve the resulting equation. We then check for any extraneous solutions by substituting back into the original equation.

Step 1: Set Up the Equation

We start with the equation given by the one-to-one property of logarithms: \[ \ln (-4x) = \ln (x^2 - 3x) \] This implies: \[ -4x = x^2 - 3x \]

Step 2: Rearrange the Equation

Rearranging the equation gives us: \[ x^2 - 3x + 4x = 0 \implies x^2 + x = 0 \] Factoring out $x$ results in: \[ x(x + 1) = 0 \]

Step 3: Solve for $x$

Setting each factor to zero, we find the solutions: \[ x = 0 \quad \text{or} \quad x = -1 \]

Step 4: Check for Extraneous Solutions

We need to check if these solutions are valid in the context of the logarithm:

For $x = 0$: $\ln(-4 \cdot 0)$ is undefined.
For $x = -1$: $\ln(-4 \cdot -1) = \ln(4)$ is valid, but $\ln((-1)^2 - 3 \cdot (-1)) = \ln(4)$ is also valid.

Thus, both solutions are extraneous.