Question

Solve: $\log (x-2)-\log (4x+16)=\log \frac{1}{x}$

Short answer

By solving the equation, $\log (x-2)-\log (4x+16)=\log \frac{1}{x}$, the value of $x$is $8$

Step-by-step solution

Step 1. Given

An expression$\log (x-2)-\log (4x+16)=\log \frac{1}{x}$

To solve the expression

Step 2. Use the Quotient Property on the left side

Use the Quotient Property on the left side

$$\begin{gathered} \log (x-2)-\log (4x+16)=\log \frac{1}{x} \\ \log \left(\frac{x-2}{4x+16}\right)=\log \frac{1}{x} \end{gathered}$$

Step 3. Use the One-to-One Property,

Step 3. Use the One-to-One Property, if ${\log }_{a}M={\log }_{a}N$, then $M=N$
$$\begin{gathered} \left(\frac{x-2}{4x+16}\right)=\frac{1}{x} \\ {x}^2-2x=4x+16 \\ {x}^2-6x-16=0 \end{gathered}$$

Step 4. Factorize the equation

$$\begin{gathered} x^2-6x-16=0 \\ (x-8)(x+2)=0 \\ x=-2,8 \end{gathered}$$

Logarithm for a negative number does not exist.

So,$x=8$

Step 5. Check the solution

Substitute $x=8$in the original equation and solve.

$$\begin{gathered} \log (8-2)-\log \left(4\right(8)+16)=\log \frac{1}{8} \\ \log \left(6\right)-\log \left(48\right)=\log \frac{1}{8} \\ \log \left(\frac{6}{48}\right)=\log \frac{1}{8} \\ \frac{1}{8}=\frac{1}{8} \end{gathered}$$

Hence the solution has been checked.