Question

Fill in the reason for each step in the following two solutions. Solve: $\log _{3}(x-1)^{2}=2$ $$ \begin{aligned} &\begin{array}{ll} {\text { Solution } A} & {\text { Solution } B} \\ {\log _{3}(x-1)^{2}=2} & {\log _{3}(x-1)^{2}=2} \end{array}\\ &(x-1)^{2}=3^{2}=9 \quad 2 \log _{3}(x-1)=2\\ &(x-1)=\pm 3 \quad \log _{3}(x-1)=1\\ &x-1=-3 \text { or } x-1=3 \quad x-1=3^{1}=3\\ &x=-2 \text { or } x=4\\ &\pi x=4 \end{aligned} $$ Both solutions given in Solution A check. Explain what caused the solution $x=-2$ to be lost in Solution B.

Name: Problem 109, Chapter 6: Algebra and Trigonometry
Uploaded: 2020-06-16T21:39:46-08:00
Duration: 2 min 35 s
Channel: James Kiss
Description: Fill in the reason for each step in the following two solutions. Solve: log3(x-1)^2=2 Solution A Solution B log3(x-1)^2=2 log3(x-1)^2=2 (x-1)^2=3^2=9 2 log3(x-1)=2 (x-1)=±3 log3(x-1)=1 x-1=-3 or x-1=3 x-1=3^1=3 x=-2 or x=4 πx=4 Both solutions given in Solution A check. Explain what caused the solution x=-2 to be lost in Solution B.

   Fill in the reason for each step in the following two solutions.
 Solve: $\log _{3}(x-1)^{2}=2$
$$
\begin{aligned}
&\begin{array}{ll}
{\text { Solution } A} & {\text { Solution } B} \\
{\log _{3}(x-1)^{2}=2} & {\log _{3}(x-1)^{2}=2}
\end{array}\\
&(x-1)^{2}=3^{2}=9 \quad 2 \log _{3}(x-1)=2\\
&(x-1)=\pm 3 \quad \log _{3}(x-1)=1\\
&x-1=-3 \text { or } x-1=3 \quad x-1=3^{1}=3\\
&x=-2 \text { or } x=4\\
&\pi x=4
\end{aligned}
$$
Both solutions given in Solution A check. Explain what caused the solution $x=-2$ to be lost in Solution B.

Algebra and Trigonometry

Michael Sullivan 10th Edition

Chapter 6, Problem 109

Instant Answer

Solved by Expert James Kiss

06/16/2020

Step 1

Step 1: We start with the given equation $\log _{3}(x-1)^{2}=2$ for both Solution A and Solution B. Show more…

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Fill in the reason for each step in the following two solutions. Solve: $\log _{3}(x-1)^{2}=2$ $$ \begin{aligned} &\begin{array}{ll} {\text { Solution } A} & {\text { Solution } B} \\ {\log _{3}(x-1)^{2}=2} & {\log _{3}(x-1)^{2}=2} \end{array}\\ &(x-1)^{2}=3^{2}=9 \quad 2 \log _{3}(x-1)=2\\ &(x-1)=\pm 3 \quad \log _{3}(x-1)=1\\ &x-1=-3 \text { or } x-1=3 \quad x-1=3^{1}=3\\ &x=-2 \text { or } x=4\\ &\pi x=4 \end{aligned} $$ Both solutions given in Solution A check. Explain what caused the solution $x=-2$ to be lost in Solution B.

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Key Concepts

Domain of Logarithmic Functions

The logarithm function is defined only for positive arguments. In the context of the problem, rewriting log?((x - 1)²) as 2 log?(x - 1) is only valid if x - 1 is positive. This restriction causes any solution where x - 1 is negative (such as x = -2) to be inadvertently excluded during manipulation.

Properties of Logarithms

When applying the property log?b?(a?) = n log?b?(a), it is critical to remember that this identity holds only when the base expression a is positive. Ignoring the absolute value that arises from taking a square causes the negative solution to be lost, as it fails to meet this condition.

Handling Squared Equations

Squaring an expression, as in (x - 1)², conceals the original sign information of x - 1, permitting both positive and negative roots when taking the square root. An appropriate solution process must consider both the positive and negative square roots to capture all valid solutions.

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