Fill in reasons for each step in the following two solutions.

Solve: ${\log }_3{\left(x-1\right)}^2=2$

Solution A

$$\begin{gathered} {\log }_3{\left(x-1\right)}^2=2 \\ {\left(x-1\right)}^2=3^2=9\_\_\_\_\_\_\_\_\_\_\_ \\ \left(x-1\right)=\pm 3\_\_\_\_\_\_\_\_\_\_\_ \\ x-1=-3orx-1=3\_\_\_\_\_\_\_\_\_\_ \\ x=-2orx=4\_\_\_\_\_\_\_\_\_\_ \end{gathered}$$

Solution B

$$\begin{gathered} {\log }_3{\left(x-1\right)}^2=2 \\ 2{\log }_3\left(x-1\right)=2\_\_\_\_\_\_\_\_\_\_ \\ {\log }_3\left(x-1\right)=1\_\_\_\_\_\_\_\_\_\_\_ \\ x-1=3^1=3\_\_\_\_\_\_\_\_\_\_ \\ x=4\_\_\_\_\_\_\_\_\_\_ \end{gathered}$$

Both solutions given in Solution A check. Explain what caused the solution$x=-2$to be lost in Solution B.

We are given two solutions A and B

$$\begin{gathered} {\log }_3{\left(x-1\right)}^2=2 \\ {\left(x-1\right)}^2=3^2=9\_\_\_\_\_\_\_\_\_\_\_ \\ \left(x-1\right)=\pm 3\_\_\_\_\_\_\_\_\_\_\_ \\ x-1=-3orx-1=3\_\_\_\_\_\_\_\_\_\_ \\ x=-2orx=4\_\_\_\_\_\_\_\_\_\_ \end{gathered}$$

and $$\begin{gathered} {\log }_3{\left(x-1\right)}^2=2 \\ 2{\log }_3\left(x-1\right)=2\_\_\_\_\_\_\_\_\_\_ \\ {\log }_3\left(x-1\right)=1\_\_\_\_\_\_\_\_\_\_\_ \\ x-1=3^1=3\_\_\_\_\_\_\_\_\_\_ \\ x=4\_\_\_\_\_\_\_\_\_\_ \end{gathered}$$

respectively.

Solution A

$$\begin{gathered} {\log }_3{\left(x-1\right)}^2=2 \\ {\left(x-1\right)}^2=3^2=9,Property:lo{g}_a\left(x\right)=y\iff x=a^y \\ \left(x-1\right)=\pm 3,x^2=a\iff \left|x\right|=\sqrt{a}\iff x=\pm \sqrt{a} \\ x-1=-3orx-1=3Becauseofthesquarerootswehavetwocases \\ x=-2orx=4SolutionsofLinearEquations \end{gathered}$$

$$\begin{gathered} {\log }_3{\left(x-1\right)}^2=2 \\ 2{\log }_3\left(x-1\right)=2,Propertylo{g}_a\left(x^r\right)=rlo{g}_a\left(x\right) \\ {\log }_3\left(x-1\right)=1,Dividethewholeequationby2 \\ x-1=3^1=3,Propertylo{g}_a\left(x\right)=y\iff x=a^y \\ x=4SolutionofLinearequation \end{gathered}$$

The functions ${\log }_3{\left(x-1\right)}^2$and $2{\log }_3\left(x-1\right)$are not equal because their domain are not equal.

The problem occurred when we put the power of $x$in front of logarithm. The argument of logarithm function was no longer a quadratic function but linear one that as only one solution.

Since the square a number is always positive, the expression$\left(x-1\right)$in${\log }_3{\left(x-1\right)}^2$ could also be a negative number . While the expression$x-1$in${\log }_3\left(x-1\right)$can not be negative.