Chapter 11, Problem 5CR

To determine

To find: the expression is ${\log }_3\left(x-1\right)+{\log }_3\left(2x+1\right)=2$ .

Answer to Problem 5CR

The expression set is $\left\{\frac{5}{2}\right\}$

Explanation of Solution

Given information:

The given equation is ${\log }_3\left(x-1\right)+{\log }_3\left(2x+1\right)=2$

Calculation:

Calculate the logarithm of a product.

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

Change the logarithm expression into an exponential expression.

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

Solve the equation fo0r $x$

The two integers whose product is -20 and sum is -1 are 4 and -5. So , $2x^2-x-10$ is equivalent to $\left(x+4\right)\left(2x-5\right)=0$

According to the zero-product property, either $x+4=0$ or $x-5=0$ or both the factors can zero.

  $\begin{array}{c}x+4=0\\ x=-4\end{array}$

  $\begin{array}{c}2x-5=0\\ x=\frac{5}{2}\end{array}$

When we substitute -4 for $x$ in the given equation, logarithm of negative numbers results is not defined. So, only the solution $x=\frac{5}{2}$ is valid.

Therefore, the solution set is $\left\{\frac{5}{2}\right\}$