Solve for the indicated variable.
-$$\ln ( x - 4 ) - \ln ( y - 1 ) = \ln 3 , \quad \text { for } x$$
A)x = y + 6
B) $x = \frac { 4 y - 1 } { y - 1 }$
C) $x = e ( y - 1 ) + 3 y + 1$
D) $x = 3 y + 1$

Question

Quizplus · Accepted Answer

Using the properties of logarithms, $\ln ( x - 4 ) - \ln ( y - 1 ) = \ln \frac{x-4}{y-1} = \ln 3$. Therefore, $\frac{x-4}{y-1} = 3$. Solving for $x$, we get $x = 3(y-1) + 4 = 3y + 1$.

Solve for the Indicated Variable ln⁡(x−4)−ln⁡(y−1)=ln⁡3, for x\ln ( x - 4 ) - \ln ( y - 1 ) = \ln 3 , \quad \text { for } xln(x−4)−ln(y−1)=ln3, for x

Solve for the Indicated Variable $\ln ( x - 4 ) - \ln ( y - 1 ) = \ln 3 , \quad \text { for } x$