Q. 101 Fill in reasons for each step in... [FREE SOLUTION]

91影视

Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 5: Q. 101 (page 317)

Fill in reasons for each step in the following two solutions.
Solve: $\log_{3} {(x - 1)}^{2} = 2$
Solution A
$\log_{3} {(x - 1)}^{2} = 2 {(x - 1)}^{2} = 3^{2} = 9_____(x - 1) = 卤 3_x - 1 = - 3 o r x - 1 = 3x = - 2 o r x = 4____$
Solution B
$\log_{3} {(x - 1)}^{2} = 2 2 \log_{3} (x - 1) = 2__________\log_{3} (x - 1) = 1_x - 1 = 3^{1} = 3x = 4__$
Both solutions given in Solution A check. Explain what caused the solution $x = - 2$ to be lost in Solution B.

Short Answer

Expert verified

Solution A

The solutions are $x = - 2 o r x = 4$

Solution B

The solution is $x = 4 .$

The solution is lost due to different domains.

Step by step solution

Step 1. Given Information

We are given two solutions A and B

$\log_{3} {(x - 1)}^{2} = 2 {(x - 1)}^{2} = 3^{2} = 9___________(x - 1) = 卤 3___________x - 1 = - 3 o r x - 1 = 3__________x = - 2 o r x = 4__________$

and $\log_{3} {(x - 1)}^{2} = 2 2 \log_{3} (x - 1) = 2__________\log_{3} (x - 1) = 1___________x - 1 = 3^{1} = 3__________x = 4__________$

respectively.

Step 2. Reason for solution A

Solution A

$\log_{3} {(x - 1)}^{2} = 2 {(x - 1)}^{2} = 3^{2} = 9, P r o p e r t y : l o g_{a} (x) = y 鈬� x = a^{y} (x - 1) = 卤 3, x^{2} = a 鈬� |x| = \sqrt{a} 鈬� x = 卤 \sqrt{a} x - 1 = - 3 o r x - 1 = 3 B e c a u s e o f t h e s q u a r e r o o t s w e h a v e t w o c a s e s x = - 2 o r x = 4 S o l u t i o n s o f L i n e a r E q u a t i o n s$

Step 2. Reasons For Solution B

$\log_{3} {(x - 1)}^{2} = 2 2 \log_{3} (x - 1) = 2, P r o p e r t y l o g_{a} (x^{r}) = r l o g_{a} (x) \log_{3} (x - 1) = 1, D i v i d e t h e w h o l e e q u a t i o n b y 2 x - 1 = 3^{1} = 3, P r o p e r t y l o g_{a} (x) = y 鈬� x = a^{y} x = 4 S o l u t i o n o f L i n e a r e q u a t i o n$

Step 4. Reason to lost solution x = -2 in Solution B.

The functions $\log_{3} {(x - 1)}^{2}$ and $2 \log_{3} (x - 1)$ 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 $(x - 1)$ in $\log_{3} {(x - 1)}^{2}$ could also be a negative number . While the expression $x - 1$ in $\log_{3} (x - 1)$ can not be negative.