Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 4 - Section 4.3 - Logarithmic Functions - 4.3 Exercises - Page 354: 105

Answer

1. The number of digits in 1000 is 4, and $log1000=3$, 2. For 10,000, the number of digits in 10,000 is 5, and $log10000=4$, 3. Any number between 1000 and 10,000 have a number of digit= $4$. 4. The common logarithm of such a number must lie between 3 and 4. 5. $m=[[log(x)]]+1$ see prove below. 6. The number $2^{100}$ have 31 digits.

Work Step by Step

1. The number of digits in 1000 is 4, and $log1000=3(=4-1)$, 2. For 10,000, the number of digits in 10,000 is 5, and $log10000=4(=5-1)$, 3. Any number between 1000 and 10,000 have a number of digit= $4$. 4. The common logarithm of such a number must lie between 3 and 4. 5. Assume the number of digits of the positive integer $x$ is $m$, based on the above observations, the common logarithm of $x$ must lie between $m-1$ and $m$, which means that $[[log(x)]]=m-1$ hence we proved that $[[log(x)]]+1=m$ the number of digits in $x$. 6. For $x=2^100$, $m=[[log(2^{100})]]+1=[[100log2]]+1=[[30.1]]+1=31$
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