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makkiz [27]
3 years ago
10

QF Q7.) Use properties of logarithms to expand the logarithmic expression as much as possible. Evaluate logarithmic expressions

without using a calculator if possible.

Mathematics
2 answers:
Lisa [10]3 years ago
8 0
\log _9\left(\sqrt[5]{\dfrac{a^6b}{81}}\right)=\dfrac{-2+\log _9\left(b\right)+6\log _9\left(a\right)}{5} 

\log _9\left(\sqrt[5]{\dfrac{a^6b}{81}}\right) =\ \textgreater \  \log _9\left(\left(\dfrac{a^6b}{81}\right)^{\dfrac{1}{5}}\right) 

\dfrac{1}{5}\log _9\left(\dfrac{a^6b}{81}\right) 

\log _9\left(\dfrac{a^6b}{81}\right)=\log _9\left(a^6b\right)-\log _9\left(81\right) 

\log _9\left(a^6b\right)-\log _9\left(81\right) 

\log _9\left(b\right)+\log _9\left(a^6\right)-2 

6\log _9\left(a\right) 

\dfrac{1}{5}\left(6\log _9\left(a\right)+\log _9\left(b\right)-2\right) 

\dfrac{1}{5}\log _9\left(b\right)+\dfrac{1}{5}\cdot \:6\log _9\left(a\right)+\dfrac{1}{5}\left(-2\right) 

\dfrac{1}{5}\log _9\left(b\right)+\dfrac{1}{5}\cdot \:6\log _9\dleft(a\right)-\frac{1}{5}\cdot \:2) 

\dfrac{1}{5}\log _9\left(b\right)+\dfrac{1}{5}\cdot \:6\log _9\left(a\right)-\dfrac{1}{5}\cdot \:2 

\dfrac{1}{5}\cdot \:6\log _9\left(a\right)=\dfrac{6}{5}\log _9\left(a\right) 

\dfrac{1}{5}\log _9\left(b\right)+\dfrac{6}{5}\log _9\left(a\right)-\dfrac{2}{5} 

\dfrac{\log _9\left(b\right)}{5}+\log _9\left(a\right)\dfrac{6}{5}-\frac{2}{5} 

\log _9\left(a\right)\dfrac{6}{5}\::\quad \dfrac{6\log _9\left(a\right)}{5} 

\dfrac{\log _9\left(b\right)}{5}+\dfrac{6\log _9\left(a\right)}{5}-\dfrac{2}{5} 

\dfrac{\log _9\left(b\right)}{5}+\dfrac{6\log _9\left(a\right)}{5}:\quad \dfrac{\log _9\left(b\right)+6\log _9\left(a\right)}{5} =\ \textgreater \  \dfrac{-2+\log _9\left(b\right)+6\log _9\left(a\right)}{5} 

\text{Took a while, but here it is. Hope it helps!}
Softa [21]3 years ago
5 0
Attached is the solution.
The following log properties are used:
log(x^n) = nlog(x)
log(\frac{a}{b}) = log(a) - log(b) \\  \\ log(ab) = log(a) + log(b) \\  \\ log_9 (81) = 2 \rightarrow 9^2 = 81



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The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

The lifetime (in hours) of a 60-watt light bulb is a random variable that has a Normal distribution with σ = 30 hours. A random sample of 25 bulbs put on test produced a sample mean lifetime of = 1038 hours.

If in the study of the lifetime of 60-watt light bulbs it was desired to have a margin of error no larger than 6 hours with 99% confidence, how many randomly selected 60-watt light bulbs should be tested to achieve this result?

Given Information:

standard deviation = σ = 30 hours

confidence level = 99%

Margin of error = 6 hours

Required Information:

sample size = n = ?

Answer:

sample size = n ≈ 165

Step-by-step explanation:

We know that margin of error is given by

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Where z is the corresponding confidence level score, σ is the standard deviation and n is the sample size

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squaring both sides

n = (z*σ/Margin of error)²

For 99% confidence level the z-score is 2.576

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