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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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Construct a​ 95% confidence interval for the population​ mean, mu. Assume the population has a normal distribution. A random sam
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Given Information:  

Number of lithium batteries = n = 16

Mean life of lithium batteries = μ = 645 hours

Standard deviation of lithium batteries = σ = 31 hours

Confidence level = 95%  

Required Information:  

Confidence Interval = ?

Answer:  

CI = 628.5 \: to \: 661.5 \: hours

Step-by-step explanation:  

The confidence interval is given by

CI = \mu \pm t_{\alpha/2} (\frac{\sigma}{\sqrt{n}})

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CI = 645 \pm 2.131(7.75})

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CI = 645 - 16.515 \: and \: 645 + 16.515

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Therefore, the 95% confidence interval is 628.5 to 661.5 hours

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It means that we are 95% confident that the mean life of 16 lithium batteries is within the interval of (628.5 to 661.5 hours)

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(9-5+X)+ (39+211)

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