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 July 29th, 2013, 02:30 PM #1 Newbie   Joined: Aug 2012 Posts: 10 Thanks: 0 How to get rid of a negative exponent on logs... Hi, I've hit a brick wall here...not sure how to proceed Some insight on the original function: ln(x)=log4(x+3)-log4(x-2) Now solve for Ln(1/2) therefore.... Ln(1/2)= Log4(1/2+3)-Log4(1/2-2) = Log4((7/2)/(-3/2)) = Log4(-7/3)....now what? i know i have to apply the natural log at this point, but why? how? i don't understand...HELP!!!
July 29th, 2013, 05:41 PM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 407

Re: How to get rid of a negative exponent on logs...

Hello, Toyboy!

Please give us the original wording of the problem.
What you presented makes little sense.

Quote:
 $\text{I've hit a brick wall here ... not sure how to proceed.}$ $\text{Some insight on the original function:} \;\;\;\ln(x)\:=\:\log_4(x\,+\,3)\,-\,\log_4(x\,-\,2)$ [color=beige] . [/color][color=blue]This can't be right![/color] $\text{Now solve for }\ln\left(\frac{1}{2}\right)$[color=beige] . [/color][color=blue]Are you SURE that's what it says?[/color] $\ln\left(\frac{1}{2}\right) \:=\: \log_4\left(\frac{1}{2}\,+\,3\right)\,-\,\log_4\left(\frac{1}{2}\,-\,2\right)$ $\ln\left(\frac{1}{2}\right) \:=\:\log_4\left(\frac{7}{2}\right)\,-\,\log_4\left(-\frac{3}{2}\right)$ $\ln\left(\frac{1}{2}\right) \:=\:\log_4\left(\frac{\frac{7}{2}}{-\frac{3}{2}}\right)$ $\ln\left(\frac{1}{2}\right) \:=\:\log_4\left(-\frac{7}{3}\right)$

$\text{You can stop . . . You've done what they've asked.}
\text{You've solved for }\,\ln\left(\frac{1}{2}\right)
\;\;\;\text{but the answer is }undefined.$

July 30th, 2013, 02:52 AM   #3
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Joined: Aug 2012

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Re: How to get rid of a negative exponent on logs...

Hi Soroban,

My apologies, attached is a pciture of the actual question....
Attached Images
 20130730_123901.jpg (48.6 KB, 160 views)

 July 30th, 2013, 03:01 AM #4 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 Re: How to get rid of a negative exponent on logs... This is what you should be doing: $0.5= Log_4(x+3)-Log_4(x-2) 0.5=log_4(\frac {x+3}{x-2}) 4^{0.5}=\frac {x+3}{x-2} 2x-4=x+3 x=7$
July 30th, 2013, 04:16 AM   #5
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Joined: Aug 2012

Posts: 10
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Re: How to get rid of a negative exponent on logs...

Quote:
 Originally Posted by gelatine1 This is what you should be doing: $0.5= Log_4(x+3)-Log_4(x-2) 0.5=log_4(\frac {x+3}{x-2}) 4^{0.5}=\frac {x+3}{x-2} 2x-4=x+3 x=7$
Thanks alot man, don't know what i was thinking....i could almost slap myself

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