Write log_3 frac{1}{27x^{2} in the form a+b log_3x where a and b are integers

Question
Logarithms
asked 2021-02-25
Write \(\log_3 \frac{1}{27x^{2}\) in the form a+b \(\log_3x\) where a and b are integers

Answers (1)

2021-02-26
Use the quotient property:
\(log_b (\frac{m}{n})=\log_b m-\log_b n\):
\(=\log_3 1-\log_3 27x^{2}\)
The logarithm of 1 (using any base) is 0:
\(=0-\log_3 27x^{2}\)
\(= -\log_3 27x^{2}\)
Use the product property: \log_b mn=\log_b m+\log_b n:
\(= -(\log_3 27+\log_3x^{2})\)
\(= -(\log_3 3^{3}+log_3x^{2})\)
Use the rule: \(\log_b b^{x}=x\):
\(= -(3+\log_3x^{2})\)
\(= -3-\log_3x^{2}\)
Use the power property: \(\log_bx^{n}=n\log_bx\)
\(= -3-2\log_3x\)
0

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