Question

In the following exercise, solve for x.

$\log x+\log (x+3)=1$

Short answer

The solution of the given expression is$x=2$

Step-by-step solution

Step 1. Given information.   

The given expression is$\log x+\log (x+3)=1$

Step 2. Use logarithmic properties       

Using product rule, we get:

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

Writing it as an exponential function

$$\begin{gathered} \log \left(x^2+3x\right)=1 \\ {x}^2+3x={10}^1 \\ {x}^2+3x-10=0 \end{gathered}$$

Step 3. Solve for x.      

Solving the equation, we get:

$$\begin{gathered} x^2+3x-10=0 \\ {x}^2+5x-2x-10=0 \\ x\left(x+5\right)-2\left(x+5\right)=0 \\ \left(x-2\right)\left(x+5\right)=0 \end{gathered}$$

$x=2$or

$x=-5$

We eliminate $x=-5$as the logarithmic function doesn't take any negative values.

Step 4. Check the value 

Substituting $x=2$in the given expression, we get:

$$\begin{gathered} \log x+\log (x+3)=1 \\ \log 2+\log \left(2+3\right)=1 \\ \log 2+\log \left(5\right)=1 \\ \log \left(2\times 5\right)=1 \\ \log 10=1 \\ 1=1 \end{gathered}$$

This is true.

Thus, $x=2$is the solution of the given function.