Question

In the following exercises, solve for x.

$\log \left(x-1\right)-\log \left(3x+5\right)=-\log x$

Short answer

The value of $x$ for the equation role="math" localid="1645038702854" $\log \left(x-1\right)-\log \left(3x+5\right)=-\log x$is $5$

Step-by-step solution

Step 1. Given data 

The given Logarithmic Equation is$\log \left(x-1\right)-\log \left(3x+5\right)=-\log x$

Step 2. Apply the quotient properties of the logarithm 

Use quotient properties of the logarithm on the left side and power property on the right side

$$\begin{gathered} \log \left(x-1\right)-\log \left(3x+5\right)=-\log x \\ \log \frac{\left(x-1\right)}{\left(3x+5\right)}=\log x^{-1} \\ \log \frac{\left(x-1\right)}{\left(3x+5\right)}=\log \frac{1}{x} \end{gathered}$$

Step 3. Apply one to one property

Use one to one property of logarithms, if ${\log }_a\left(m\right)={\log }_a\left(n\right)$

then $m=n$

so$$\begin{gathered} \log \frac{\left(x-1\right)}{\left(3x+5\right)}=\log \frac{1}{x} \\ \frac{\left(x-1\right)}{\left(3x+5\right)}=\frac{1}{x} \\ x(x-1)=(3x+5) \\ {x}^2-x=3x+5 \\ {x}^2-4x-5=0 \\ (x+1)(x-5)=0 \end{gathered}$$

so$x=5,-1$

the logarithm of a negative number is not possible

so$x=5$

Step 4. verification  

Substitute $5$ for x in the equation

$$\begin{gathered} \log \left(x-1\right)-\log \left(3x+5\right)=-\log x \\ \log \left(5-1\right)-\log \left(3\left(5\right)+5\right)\stackrel{?}{=}-\log 5 \\ \log 4-\log \left(15+5\right)\stackrel{?}{=}\log 5^{-1} \\ \log 4-\log 20\stackrel{?}{=}\log \frac{1}{5} \\ \log \left(\frac{4}{20}\right)\stackrel{?}{=}\log \frac{1}{5} \\ \log \frac{1}{5}=\log \frac{1}{5} \end{gathered}$$