Q. 5 - Precalculus Enhanced with Graphing Utilities

Question

In Problems $1-6$, solve each equation.

${\log }_3\left(x-1\right)+{\log }_3\left(2x+1\right)=2$.

Step-by-Step Solution

Answer

The solution for the equation ${\log }_3\left(x-1\right)+{\log }_3\left(2x+1\right)=2$ is, $x=\frac{5}{2}$.

Step 1 Given equation is

${\log }_3\left(x-1\right)+{\log }_3\left(2x+1\right)=2$.

The logarithmic property used in this problem is,

$\log \left(m\right)+\log \left(n\right)=\log \left(mn\right)$

Now apply this property in the given equation.

${\log }_3\left((x-1)\left(2x+1\right)\right)=2$.

Step 2 Change the logarithmic expression into an exponential expression and simplify it.

$$\begin{gathered} \left(x-1\right)\left(2x+1\right)={\left(3\right)}^2 \\ 2x^2+x-2x-1=9 \\ 2x^2-x-1=9 \\ 2x^2-x-10=0 \end{gathered}$$

Now factorize the obtained equation.

$\left(x+4\right)\left(2x-5\right)=0$.

Step 3 Now use the zero product property and find the values for x .

$$\begin{gathered} \left(x+4\right)=0 or \left(2x-5\right)=0 \\ x=-4 or 2x=5 \\ x=\frac{5}{2} \end{gathered}$$

Since the logarithm of negative numbers cannot be defined, $x=-4$ is not valid.

So, the solution is, $x=\frac{5}{2}$.