## Precalculus (6th Edition) Blitzer

The required value is $x=2$
Consider the given equation, $\ln \left( x+4 \right)-\ln \left( x+1 \right)=\ln x$ Apply the quotient rule of logarithms \begin{align} & \ln \left( x+4 \right)-\ln \left( x+1 \right)=\ln x \\ & \ln \frac{\left( x+4 \right)}{\left( x+1 \right)}=\ln x \end{align} Comparing both sides, we get, \begin{align} & \frac{\left( x+4 \right)}{\left( x+1 \right)}=x \\ & \left( x+4 \right)=x\left( x+1 \right) \\ & \left( x+4 \right)={{x}^{2}}+x \\ \end{align} Add to both sides $-x$ \begin{align} & \left( x+4-x \right)={{x}^{2}}+x-x \\ & {{x}^{2}}=4 \end{align} Taking the square root on both sides, we get: \begin{align} & \sqrt{{{x}^{2}}}=\sqrt{4} \\ & x=2,-2 \end{align} Here $x=-2$ is not the solution of the given expression because \begin{align} & \ln \left( x+4 \right)-\ln \left( x+1 \right)=\ln x \\ & \ln \left( -2+4 \right)-\ln \left( -2+1 \right)=\ln \left( -2 \right) \\ & \ln \left( 2 \right)-\ln \left( -1 \right)=\ln \left( -2 \right) \end{align} So, by definition of ${{\log }_{a}}b$, $b>0$ but $-1$ and $-2$ is not greater than zero Thus, $x=-2$ does not satisfy the equation $\ln \left( x+4 \right)-\ln \left( x+1 \right)=\ln x$ Hence, $x=2$ is the only solution of the given expression. Therefore, the value of $\ln \left( x+4 \right)-\ln \left( x+1 \right)=\ln x$ is $x=2$.