100M Laura was asked to expand the expression log in terms of log M, log N, and constants. Laura's work/explanation: First I have to use the 2nd Log Law to turn the log of a quotient into a difference of logs: 100M log = log(100M) – log N² Now I have to use the 1st Log Law to turn the log of a product into a sum of logs: log 100 + log M - log N2 And finally I have to use the 3rd Log Law to turn the argument's exponent into the log's coefficient: log 100 + log M- 2 log N
100M Laura was asked to expand the expression log in terms of log M, log N, and constants. Laura's work/explanation: First I have to use the 2nd Log Law to turn the log of a quotient into a difference of logs: 100M log = log(100M) – log N² Now I have to use the 1st Log Law to turn the log of a product into a sum of logs: log 100 + log M - log N2 And finally I have to use the 3rd Log Law to turn the argument's exponent into the log's coefficient: log 100 + log M- 2 log N
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![100M
Laura was asked to expand the expression log
in terms of log M, log N, and constants.
Laura's work/explanation:
First I have to use the 2nd Log Law to turn the log of a
quotient into a difference of logs:
100M
log
= log(100M) – log N2
Now I have to use the 1st Log Law to turn the log of a
product into a sum of logs:
log 100 + log M – log N?
And finally I have to use the 3rd Log Law to turn the
argument's exponent into the log's coefficient:
log 100 + log M – 2 log N](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F10a9f489-d438-4273-b76d-cc8abf92d638%2Fda6679bc-5bdb-4a54-844b-f91ef1c7e9bf%2F9nhzm2s_processed.jpeg&w=3840&q=75)
Transcribed Image Text:100M
Laura was asked to expand the expression log
in terms of log M, log N, and constants.
Laura's work/explanation:
First I have to use the 2nd Log Law to turn the log of a
quotient into a difference of logs:
100M
log
= log(100M) – log N2
Now I have to use the 1st Log Law to turn the log of a
product into a sum of logs:
log 100 + log M – log N?
And finally I have to use the 3rd Log Law to turn the
argument's exponent into the log's coefficient:
log 100 + log M – 2 log N
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