Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 9 - Section 9.5 - Exponential and Logarithmic Equations - Exercise Set - Page 727: 95

Answer

The solution set is $\{12\}$.

Work Step by Step

The given equation is $\log_2(x-6)+\log_2(x-4)-\log_2x=2$ Use product rule. $\log_2(x-6)(x-4)-\log_2x=2$ Use quotient rule. $\log_2 \frac{(x-6)(x-4)}{x}=2$ Write the equation in exponent form. $ \frac{(x-6)(x-4)}{x}=2^2$ $ \frac{(x-6)(x-4)}{x}=4$ Multiply both sides by $x$ $x\cdot \frac{(x-6)(x-4)}{x}=x\cdot 4$ Simplify. $(x-6)(x-4)= 4x$ Use the distributive property on the left hand side. $ x^2 -6x-4x+24= 4x$ Subtract $4x$ from both sides. $ x^2 -6x-4x+24-4x= 4x-4x$ Simplify. $ x^2 -14x+24= 0$ Factor. $ x^2 -12x-2x+24= 0$ $ x(x -12)-2(x-12)= 0$ $ (x -12)(x-2)= 0$ Set each factor equal to zero. $ x -12=0 $ or $x-2= 0$ Isolate $x$. $ x =12 $ or $x=2$ Check the solution. For $x=12$ $\log_2(12-6)+\log_2(12-4)-\log_212=2$ $\log_2(6)+\log_2(8)-\log_212=2$ $\log_2(6\cdot8)-\log_212=2$ $\log_2\left (\frac{6\cdot8}{12} \right )=2$ $\log_2\left (4 \right )=2$ $\log_2\left (2^2 \right )=2$ $2=2$ True. For $x=2$ $\log_2(2-6)+\log_2(2-4)-\log_22=2$ $\log_2(-4)+\log_2(-2)-\log_22=2$ Not a solution.
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