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 November 3rd, 2011, 03:27 PM #1 Newbie   Joined: Sep 2011 Posts: 13 Thanks: 0 solving logarithmic equations The problem is (x^2)e^4x + (7x)e^4x - e^4x I took e^4x out and I'm left with e^4x(x^2+7x-1) How do I solve from here? I can't figure out how to factor to get my smaller and larger values.
 November 3rd, 2011, 04:02 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,753 Thanks: 2136 If you were asked to factor the expression, you've already done enough. As there are no logarithms in the expression, why does the topic title refer to logarithmic equations?
 November 3rd, 2011, 04:12 PM #3 Newbie   Joined: Sep 2011 Posts: 13 Thanks: 0 Re: solving logarithmic equations Because that is the section of the chapter we are in. We are supposed to use logs to solve I think. We have to have two answers: a smaller value and larger value.
November 3rd, 2011, 05:00 PM   #4
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Re: solving logarithmic equations

Quote:
 Originally Posted by hannah2329 The problem is (x^2)e^4x + (7x)e^4x - e^4x
But that's NOT an equation.

 November 3rd, 2011, 05:20 PM #5 Newbie   Joined: Sep 2011 Posts: 13 Thanks: 0 Re: solving logarithmic equations It is supposed to be equal to zero. Sorry, I forgot to put it.
 November 3rd, 2011, 11:42 PM #6 Senior Member   Joined: Aug 2011 Posts: 334 Thanks: 8 Re: solving logarithmic equations The solutions of an equation of the form : f(x) g(x) = 0 are the solutions of f(x) =0 and the solutions of g(x) = 0 insofar as solution(s) exist.
 November 4th, 2011, 01:38 AM #7 Senior Member   Joined: Oct 2011 From: India ???? Posts: 224 Thanks: 0 Re: solving logarithmic equations It must be noted that $e^{4x}=0$ has no solutions. $x^2.e^{4x}+7x.e^{4x}-e^{4x}=0 \\e^{4x}(x^2+7x-1)=0\\(x^2+7x-1)=0\\x=??!$ Use quadratic formula.

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