Question

Solve: $\log (x+2)-\log (4x+3)=-\log x$

Short answer

By solving the equation, $\log (x+2)-\log (4x+3)=-\log x$, the value of $x$is $3$

Step-by-step solution

Step 1. Given

An expression$\log (x+2)-\log (4x+3)=-\log x$

To solve the expression

Step 2. Use the Quotient Property on the left side and the Power Property on the right.

Use the Quotient Property on the left side and the Power Property on the right.

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

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

Use the One-to-One Property, if ${\log }_{a}M={\log }_{a}N$, then $M=N$

$$\begin{gathered} \frac{x+2}{4x+3}=\frac{1}{x} \\ {x}^2+2x=4x+3 \\ {x}^2-2x-3=0 \end{gathered}$$

Step 4. Factorize the equation

$$\begin{gathered} x^2-2x-3=0 \\ (x-3)(x+1)=0 \\ x=-1,3 \end{gathered}$$

Logarithm for a negative number does not exist.

So,$x=3$

Step 5. Check the solution

Substitute $x=3$in the original equation and solve.

$$\begin{gathered} \log (3+2)-\log \left(4\right(3)+3)=-\log 3 \\ \log \left(5\right)-\log \left(15\right)=\log \frac{1}{3} \\ \log \left(\frac{5}{15}\right)=\log \frac{1}{3} \\ \frac{1}{3}=\frac{1}{3} \end{gathered}$$

Hence the solution has been checked.