Two questions, need help. (1 Viewer)

BlueGas · Oct 8, 2015

I need help for these two questions.
They're two separate questions by the way.

rand_althor · Oct 8, 2015

First one requires you to manipulate the terms and use log laws to get log₁₀(2) in terms of log₅(8):

$\begin{align*}\log_5{8}&=a \\\log_{10}{2}&=\frac{\log_5{2}}{\log_5{10}} \\&=\frac{\log_5{2}}{\log_5{5}+\log_5{2}} \\&=\frac{\frac{1}{3}\times3\times\log_5{2}}{1+\frac{1}{3}\times3\times\log_5{2}} \\&=\frac{\frac{1}{3}\log_5{8}}{1+\frac{1}{3}\log_5{8}} \\&=\frac{\frac{1}{3}a}{1+\frac{1}{3}a} \\&=\frac{\frac{a}{3}}{\frac{a+3}{3}} \\&=\frac{a}{a+3}\end{align*}$

For the second one you have to find the second derivative and then prove that it is always less than zero.

BlueGas · Oct 8, 2015

rand_althor said:
First one requires you to manipulate the terms and use log laws to get log₁₀(2) in terms of log₅(8):

$\begin{align*}\log_5{8}&=a \\\log_{10}{2}&=\frac{\log_5{2}}{\log_5{10}} \\&=\frac{\log_5{2}}{\log_5{5}+\log_5{2}} \\&=\frac{\frac{1}{3}\times3\times\log_5{2}}{1+\frac{1}{3}\times3\times\log_5{2}} \\&=\frac{\frac{1}{3}\log_5{8}}{1+\frac{1}{3}\log_5{8}} \\&=\frac{\frac{1}{3}a}{1+\frac{1}{3}a} \\&=\frac{\frac{a}{3}}{\frac{a+3}{3}} \\&=\frac{a}{a+3}\end{align*}$

For the second one you have to find the second derivative and then prove that it is always less than zero.

How did you get the first step in log5(2)? I thought a = log5(8)?

BlueGas · Oct 8, 2015

I mean how did you get log5?

InteGrand · Oct 8, 2015

BlueGas said:
How did you get the first step in log5(2)? I thought a = log5(8)?

The first step is just the change of base rule.

rand_althor · Oct 8, 2015

BlueGas said:
I mean how did you get log5?

The first and second lines are not linked. I was just stating what 'a' was equal to. The change of base rule is: $\log_a{b}=\frac{\log_m{a}}{\log_m{b}}$

Did you get the second question?

BlueGas · Oct 8, 2015

rand_althor said:
The first and second lines are not linked. I was just stating what 'a' was equal to. The change of base rule is: $\log_a{b}=\frac{\log_m{a}}{\log_m{b}}$

Did you get the second question?

Yeah I know how to do the second question but the first question is waaaay too confusing.

Two questions, need help. (1 Viewer)

BlueGas

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rand_althor

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BlueGas

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BlueGas

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InteGrand

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rand_althor

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BlueGas

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