Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Section 4.5 - Exponential and Logarithmic Functions - 4.5 Exercises - Page 368: 15

Answer

$x=-\dfrac{\log3-\log2}{\log2}\approx-0.584963$

Work Step by Step

$2^{1-x}=3$ We can write the left side of the equation like this: $(2)(2^{-x})=3$ Take the $2$ to divide the right side of the equation. It becomes: $2^{-x}=\dfrac{3}{2}$ Apply $\log$ to both sides of the equation: $\log2^{-x}=\log{\dfrac{3}{2}}$ Now, the exponent $-x$ can be taken down to multiply in front of the logarithm: $-x\log2=\log{\dfrac{3}{2}}$ Solve for $x$: $-x=\dfrac{\log{\dfrac{3}{2}}}{\log2}$ $x=-\dfrac{\log{\dfrac{3}{2}}}{\log2}$ We know that the $\log$ of a division can be expanded as a substraction, so we can write the numerator like this: $x=-\dfrac{\log3-\log2}{\log2}\approx-0.584963$
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