Question

In the following exercises, solve for $x$

$\log (x+4)-\log (5x+12)=-\log x$

Short answer

By solving the equation $\log (x+4)-\log (5x+12)=-\log x$, the value of $x$is $4$

Step-by-step solution

Step 1. Given

An expression$\log (x+4)-\log (5x+12)=-\log x$

To solve the expression for $x$

Step 2. Use the Quotient property

By Quotient Property of logarithm,

${\log }_{a}M-{\log }_{a}N={\log }_a\frac{M}{N}$

$$\begin{gathered} \log (x+4)-\log (5x+12)=-\log x \\ \log \left(\frac{x+4}{5x+12}\right)=\log x^{-1} \\ \log \left(\frac{x+4}{5x+12}\right)=\log \frac{1}{x} \end{gathered}$$

Step 3. Use One to One property

If ${\log }_{a}M={\log }_{a}N$, then $M=N$.

So, $\frac{x+4}{5x+12}=\frac{1}{x}$

Step 4. Solve the equation

$\frac{x+4}{5x+12}=\frac{1}{x}$

$$\begin{gathered} x^2-x-12=0 \\ {x}^2-4x+3x-12=0 \\ (x-4)(x+3)=0 \\ x=-3,4 \end{gathered}$$

Logarithm for a negative number does not exist.

So, $x=4$

Step 5. Check the solution

Substitute $x=4$in the original equation and solve.

$$\begin{gathered} \log (4+4)-\log \left(5\right(4)+12)=-\log \left(4\right) \\ \log 8-\log 32=\log 4^{-1} \\ \log \frac{8}{32}=\log \frac{1}{4} \\ \log \frac{1}{4}=\log \frac{1}{4} \end{gathered}$$

Hence the solution is checked.