Approximate In (4.03) (5.02)².
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![The given text in the image states:
"Approximate ln(4.03)(5.02)^2."
This text is asking to find the natural logarithm (ln) of the product of 4.03 and the square of 5.02. The natural logarithm is the logarithm to the base e (where \( e \approx 2.71828 \)).
This problem involves using logarithm rules to simplify and calculate the expression. Remember that for any positive real numbers \(a\) and \(b\) and any real number \(c\):
\[ \ln(ab) = \ln(a) + \ln(b) \]
\[ \ln(a^c) = c \cdot \ln(a) \]
So, to solve \( \ln(4.03 \cdot 5.02^2) \), we can break it down as follows:
\[
\ln(4.03 \cdot (5.02)^2) = \ln(4.03) + \ln((5.02)^2) = \ln(4.03) + 2 \cdot \ln(5.02)
\]
Using these rules, you can approximate the logarithmic values using a calculator or logarithm tables.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa40558bc-afda-4e2b-be1a-c4211c27f73c%2F6e733ec2-7295-4f53-9fd8-0cb7e6ec69a3%2Fry76p6y_processed.png&w=3840&q=75)
Transcribed Image Text:The given text in the image states:
"Approximate ln(4.03)(5.02)^2."
This text is asking to find the natural logarithm (ln) of the product of 4.03 and the square of 5.02. The natural logarithm is the logarithm to the base e (where \( e \approx 2.71828 \)).
This problem involves using logarithm rules to simplify and calculate the expression. Remember that for any positive real numbers \(a\) and \(b\) and any real number \(c\):
\[ \ln(ab) = \ln(a) + \ln(b) \]
\[ \ln(a^c) = c \cdot \ln(a) \]
So, to solve \( \ln(4.03 \cdot 5.02^2) \), we can break it down as follows:
\[
\ln(4.03 \cdot (5.02)^2) = \ln(4.03) + \ln((5.02)^2) = \ln(4.03) + 2 \cdot \ln(5.02)
\]
Using these rules, you can approximate the logarithmic values using a calculator or logarithm tables.
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