Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Diagnostic Tests - A Diagnostic Tests: Algebra: 1

Answer

1. Evaluate each expression without using a calculator. a) 81 b)-81 c) $\frac{1}{81}$ d)25 e)$\frac{9}{4}$ f) $\frac{1}{8}$

Work Step by Step

1. Evaluate each expression without using a calculator. (a) $(-3)^{4}$ Answer: Using: minus one times minus one equals one. ( $-1\times-1 = +1 $) and having $a^{n} = a\times a\times...\times a_{n}$ then: $ (-3)\times(-3)\times (-3)\times(-3)$ = 81 (b) $-3^{4}$ Answer: $-3^{4} = - (3^{4})$ having $a^{n} = a\times a\times...\times a_{n}$ : $- ( 3\times3\times3\times3)$ = -81 c) $3^{(-4) }$ Answer: Having: $a^{-1} = \frac{1}{a}$ $\frac{1}{(3\times3\times3\times3)}$= $\frac{1}{81}$ d) $\frac{5^{23}}{5^{21}}$ Answer : Consider: $\frac{a^{b}}{a^{c}} = a^{b-c} $ $5^{2}$= 25 e) $(\frac{2}{3})^{-2}$ Answer: Having: $(\frac{a}{b})^{-1} = \frac{b}{a}$ and considering $(\frac{b}{a})^{n}= \frac{b^{n}}{a^{n}}$ $(\frac{2}{3})^{-2} = \frac{3\times3}{2\times2}$ $\frac{9}{4}$ f)$ 16^{-3/4} $ Answer: Consider $a^{\frac{b}{c}}= \sqrt[c] (a^{b})$ Having $ \sqrt[4] 16^{- 3} = \sqrt[4] \frac{1}{16\times16\times16} = \sqrt[4] (\frac{1}{16}\times\frac{1}{16}\times\frac{1}{16})$ If we decompose the number 16 we can find the 4th square by finding a number that multipied 4 times is 16. Then: $16\div2 = 8 $ $8\div2 = 4 $ $4\div2= 2 $ Then we know that $2\times4 = 16 $and that$ \sqrt[4] 16 = 2 $ Considering the expression $\sqrt[a] (\frac{b}{c}) =\frac{\sqrt[a] b}{\sqrt[a] c}$ and having $\sqrt[k] 1 = 1$ (for any k) $ \sqrt[4] \frac{1}{16\times16\times16} = \sqrt[4] (\frac{1}{16}\times\frac{1}{16}\times\frac{1}{16}) = \frac{1}{2}\times\frac{1}{2}\times\frac{1}{2} = \frac{1}{8}$
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