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March 26th, 2012, 02:35 PM   #1
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Show the statement is true

How would you show this statement is true? I tried finding a common denominator for the left side then subtract, but I don't know what to do from there, I'm completely stumped, can anyone help me out please?
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 March 26th, 2012, 02:47 PM #2 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Show the statement is true There's probably a slicker way (and I can think of who might know it...!), but I would use the change of base formula to rewrite the denominators as ln(a)/ln(2) ln(a)/[2ln(2)], and -ln(a)/ln(2) Then do some algebra...
March 26th, 2012, 03:19 PM   #3
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Re: Show the statement is true

Quote:
 Originally Posted by The Chaz There's probably a slicker way (and I can think of who might know it...!), but I would use the change of base formula to rewrite the denominators as ln(a)/ln(2) ln(a)/[2ln(2)], and -ln(a)/ln(2) Then do some algebra...
I don't follow, can you please explain it a bit simpler =X

 March 26th, 2012, 03:21 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,193 Thanks: 504 Math Focus: Calculus/ODEs Re: Show the statement is true We are given: $\frac{3}{\log_2(a)}-\frac{2}{\log_4(a)}=\frac{1}{\log_{\small{\frac{1} {2}}}(a)}$ Use the identity $\log_b(a)=\frac{1}{\log_a(b)}$ to write: $3\log_a(2)-2\log_a$$2^2$$=\log_a$$\frac{1}{2}$$$ $$$3-4$$\log_a(2)=-\log_a$$2$$$ $-\log_a$$2$$=-\log_a$$2$$$ True.
 March 26th, 2012, 03:24 PM #5 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Show the statement is true THERE we go!
March 26th, 2012, 03:28 PM   #6
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Re: Show the statement is true

Quote:
 Originally Posted by MarkFL We are given: $\frac{3}{\log_2(a)}-\frac{2}{\log_4(a)}=\frac{1}{\log_{\small{\frac{1} {2}}}(a)}$ Use the identity $\log_b(a)=\frac{1}{\log_a(b)}$ to write: $3\log_a(2)-2\log_a$$2^2$$=\log_a$$\frac{1}{2}$$$ $$$3-4$$\log_a(2)=-\log_a$$2$$$ $-\log_a$$2$$=-\log_a$$2$$$ True.

How does the right side become a -?

 March 26th, 2012, 03:35 PM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,193 Thanks: 504 Math Focus: Calculus/ODEs Re: Show the statement is true $\log_a$$\frac{1}{2}$$=\log_a$$2^{\small{-1}}$$=-\log_a(2)$
March 26th, 2012, 03:37 PM   #8
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Re: Show the statement is true

Quote:
 Originally Posted by MarkFL $\log_a$$\frac{1}{2}$$=\log_a$$2^{\small{-1}}$$=-\log_a(2)$

ahh I see, thank you very much! I really appreciate it!

March 26th, 2012, 03:42 PM   #9
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Re: Show the statement is true

Hello, Lucida!

I tried a "slicker" method and it turned out to be long and very involved.

Quote:
 $\text{7. Show that the following statement is true::}$ [color=beige]. . [/color]$\frac{3}{\log_2(a)}\,-\,\frac{2}{\log_4(a)} \;=\;\frac{1}{\log_{\frac{1}{2}}(a)}$

$\text{Let }\log_4(a) \,=\,P \;\;\;\Rightarrow\;\;\;4^P \,=\,a \;\;\;\Rightarrow\;\;\;(2^2)^P \,=\,a \;\;\;\Rightarrow\;\;\;2^{2P} \,=\,a$

$\text{Take logs, base 2: }\:\log_2(2^{2P}) \:=\:\log_2(a) \;\;\;\Rightarrow\;\;\;2P\cdot\underbrace{\log_2(2 )}_{\text{This is 1}} \:=\:\log_2(a)$
[color=beige]. . [/color]$2P \:=\:\log_2(a) \;\;\;\Rightarrow\;\;\;P \:=\:\frac{1}{2}\,\!\log_2(a)$

$\text{Hence: }\:\log_4(a) \:=\:\frac{1}{2}\,\!\log_2(a)$

$\text{The left side becomes: }\:\frac{3}{\log_2(a)} \,-\,\frac{2}{\frac{1}{2}\,\!\log_2(a)} \;=\;\frac{3}{\log_2(a)}\,-\,\frac{4}{\log_2(a)} \;=\;-\,\frac{1}{\log_2(a)} \;\;[1]$

$\text{Let }\log_2(a) \,=\,Q \;\;\;\Rightarrow\;\;\;2^Q \,=\,a \;\;\;\Rightarrow\;\;\;\frac{1}{2^Q} \,=\,\frac{1}{a} \;\;\;\Rightarrow\;\;\;\left(\frac{1}{2}\right)^Q \,=\,a^{-1}$

$\text{Take logs, base }\frac{1}{2}:\;\;\log_{\frac{1}{2}}\left(\frac{1}{ 2}\right)^Q \:=\:\log_{\frac{1}{2}}(a^{-1}) \;\;\;\Rightarrow\;\;\;Q\cdot\underbrace{\log_{\fr ac{1}{2}}\left(\frac{1}{2}\right)}_{\text{This is 1}} \:=\:-\,\log_{\frac{1}{2}}(a)$
$\text{Hence: }\:Q \:=\:-\log_{\frac{1}{2}}(a) \;\;\;\Rightarrow\;\;\;\log_2(a) \:=\:-\log_{\frac{1}{2}}(a)$

$\text{Substitute into [1].}$

$\text{The left side becomes: }\:-\,\frac{1}{\log_2(a)} \;=\;-\,\frac{1}{-\log_{\frac{1}{2}}(a)} \;=\;\frac{1}{\log_{\frac{1}{2}}(a)}$

Edit: Wow! . . . Lovely work, Mark!
[color=beige] .[/color]

March 26th, 2012, 04:08 PM   #10
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Re: Show the statement is true

Quote:
Originally Posted by soroban
Hello, Lucida!

I tried a "slicker" method and it turned out to be long and very involved.

Quote:
 $\text{7. Show that the following statement is true::}$ [color=beige]. . [/color]$\frac{3}{\log_2(a)}\,-\,\frac{2}{\log_4(a)} \;=\;\frac{1}{\log_{\frac{1}{2}}(a)}$

$\text{Let }\log_4(a) \,=\,P \;\;\;\Rightarrow\;\;\;4^P \,=\,a \;\;\;\Rightarrow\;\;\;(2^2)^P \,=\,a \;\;\;\Rightarrow\;\;\;2^{2P} \,=\,a$

$\text{Take logs, base 2: }\:\log_2(2^{2P}) \:=\:\log_2(a) \;\;\;\Rightarrow\;\;\;2P\cdot\underbrace{\log_2(2 )}_{\text{This is 1}} \:=\:\log_2(a)$
[color=beige]. . [/color]$2P \:=\:\log_2(a) \;\;\;\Rightarrow\;\;\;P \:=\:\frac{1}{2}\,\!\log_2(a)$

$\text{Hence: }\:\log_4(a) \:=\:\frac{1}{2}\,\!\log_2(a)$
Mark's method is a lot shorter and easier to understand, but thank you anyways!

Quote:
 $\text{The left side becomes: }\:\frac{3}{\log_2(a)} \,-\,\frac{2}{\frac{1}{2}\,\!\log_2(a)} \;=\;\frac{3}{\log_2(a)}\,-\,\frac{4}{\log_2(a)} \;=\;-\,\frac{1}{\log_2(a)} \;\;[1]$ $\text{Let }\log_2(a) \,=\,Q \;\;\;\Rightarrow\;\;\;2^Q \,=\,a \;\;\;\Rightarrow\;\;\;\frac{1}{2^Q} \,=\,\frac{1}{a} \;\;\;\Rightarrow\;\;\;\left(\frac{1}{2}\right)^Q \,=\,a^{-1}$ $\text{Take logs, base }\frac{1}{2}:\;\;\log_{\frac{1}{2}}\left(\frac{1}{ 2}\right)^Q \:=\:\log_{\frac{1}{2}}(a^{-1}) \;\;\;\Rightarrow\;\;\;Q\cdot\underbrace{\log_{\fr ac{1}{2}}\left(\frac{1}{2}\right)}_{\text{This is 1}} \:=\:-\,\log_{\frac{1}{2}}(a)$ $\text{Hence: }\:Q \:=\:-\log_{\frac{1}{2}}(a) \;\;\;\Rightarrow\;\;\;\log_2(a) \:=\:-\log_{\frac{1}{2}}(a)$ $\text{Substitute into [1].}$ $\text{The left side becomes: }\:-\,\frac{1}{\log_2(a)} \;=\;-\,\frac{1}{-\log_{\frac{1}{2}}(a)} \;=\;\frac{1}{\log_{\frac{1}{2}}(a)}$ Edit: Wow! . . . Lovely work, Mark! [color=beige] .[/color]

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