Argh... what am I not seeing? (homework problem)

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Aviatrix
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Argh... what am I not seeing? (homework problem)

Postby Aviatrix » Thu Jan 22, 2009 9:08 am UTC

I should know this stuff. I took a semester with no math courses and used those 18 weeks to forget everything I've ever known, apparently. I beat into submission most of the problems on the review of pre-calculus but this one is eluding me somehow.

Problem: Solve 6e-x - ex = 1

So I did this:

ln(6e-x) - ln(ex) = ln(1)
ln(6) + ln(e-x) - ln(ex) = ln(1)
ln(6) - x - x = 0
2x = ln(6)
x = ln(6) / 2

But when I plug that answer into my calculator, it comes up 0, not 1. I'm missing something and I've been beating my head against the wall which isn't helping. It may come to me in the middle of the night but I doubt it; exponential functions was the point in my summer pre-calculus class where I got overwhelmed with the pace of the summer session, and it's coming back to haunt me now.

Thanks for any direction.

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Re: Argh... what am I not seeing? (homework problem)

Postby The Pathological Case » Thu Jan 22, 2009 9:19 am UTC

Aviatrix wrote:I should know this stuff. I took a semester with no math courses and used those 18 weeks to forget everything I've ever known, apparently. I beat into submission most of the problems on the review of pre-calculus but this one is eluding me somehow.

Problem: Solve 6e-x - ex = 1

So I did this:

ln(6e-x) - ln(ex) = ln(1)
ln(6) + ln(e-x) - ln(ex) = ln(1)
ln(6) - x - x = 0
2x = ln(6)
x = ln(6) / 2

But when I plug that answer into my calculator, it comes up 0, not 1. I'm missing something and I've been beating my head against the wall which isn't helping. It may come to me in the middle of the night but I doubt it; exponential functions was the point in my summer pre-calculus class where I got overwhelmed with the pace of the summer session, and it's coming back to haunt me now.

Thanks for any direction.


The problem is this step
ln(6e-x - ex) = ln(1)
ln(6e-x) - ln(ex) = ln(1)

Logarithms don't distribute over sums. ln(a + b) =/= ln(a) + ln(b)
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Aviatrix
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Re: Argh... what am I not seeing? (homework problem)

Postby Aviatrix » Thu Jan 22, 2009 9:33 am UTC

Hmmm... good to know, and thanks. Unfortunately, that takes me back to where I have no idea how to approach the problem.

I went over all of the exponential/logarithm problems we did in precalculus and we never had one with more than one term unless it was a quadratic equation. That doesn't mean I shouldn't be able to figure this out, but at the moment I can't. (And I'm going to sleep; it will be here in the morning!)

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Re: Argh... what am I not seeing? (homework problem)

Postby Token » Thu Jan 22, 2009 9:42 am UTC

Change of variables: y=ex, solve for y as a quadratic, then take log for x.
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Re: Argh... what am I not seeing? (homework problem)

Postby jestingrabbit » Thu Jan 22, 2009 11:50 am UTC

Token wrote:Change of variables: y=ex, solve for y as a quadratic, then take log for x.


As a quadratic?
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Re: Argh... what am I not seeing? (homework problem)

Postby jjono » Thu Jan 22, 2009 11:55 am UTC

jestingrabbit wrote:As a quadratic?


If y=e^x then e^(-x)=1/y.

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Re: Argh... what am I not seeing? (homework problem)

Postby jestingrabbit » Thu Jan 22, 2009 2:45 pm UTC

I was seeing both exponents as x... apologies for the interuption.
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Re: Argh... what am I not seeing? (homework problem)

Postby gnuoym » Thu Jan 22, 2009 2:57 pm UTC

Token wrote:Change of variables: y=ex, solve for y as a quadratic, then take log for x.


Spoiler:
x = ln(2)?


Since ex is never zero I multiplied through the equation by it, then did the u substitution and solved the quadratic. Double checked on graphing calculator.

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Re: Argh... what am I not seeing? (homework problem)

Postby joeframbach » Thu Jan 22, 2009 3:05 pm UTC

gnuoym wrote:
Token wrote:Change of variables: y=ex, solve for y as a quadratic, then take log for x.


Spoiler:
x = ln(2)?


Since ex is never zero I multiplied through the equation by it, then did the u substitution and solved the quadratic. Double checked on graphing calculator.

Ditto on that. y=ex is the trick.

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Aviatrix
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Re: Argh... what am I not seeing? (homework problem)

Postby Aviatrix » Thu Jan 22, 2009 4:54 pm UTC

Believe it or not, even with those broad hints it took me a while, as any skill I had with algebra flew out the window this morning. But it came back, mocking me as it did.

One final question: do I say that ln^e = -3 is "illegal", "undefined", "nonsensical", or just draw a big ol' X through it? (Terminology question, not a math question, so technically off topic.)

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Re: Argh... what am I not seeing? (homework problem)

Postby mordacil » Thu Jan 22, 2009 5:28 pm UTC

One final question: do I say that ln^e = -3 is "illegal", "undefined", "nonsensical", or just draw a big ol' X through it? (Terminology question, not a math question, so technically off topic.)

[imath]ln^e[/imath] doesn't exist

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Re: Argh... what am I not seeing? (homework problem)

Postby parallax » Thu Jan 22, 2009 7:02 pm UTC

ln(x) [imath]\neq[/imath] -3 for any x
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Re: Argh... what am I not seeing? (homework problem)

Postby Buttons » Thu Jan 22, 2009 7:09 pm UTC

Except for e-3, of course.

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Re: Argh... what am I not seeing? (homework problem)

Postby parallax » Thu Jan 22, 2009 7:25 pm UTC

Sorry, I meant ex [imath]\neq[/imath] -3.
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Re: Argh... what am I not seeing? (homework problem)

Postby You, sir, name? » Thu Jan 22, 2009 7:34 pm UTC

parallax wrote:Sorry, I meant ex [imath]\neq[/imath] -3.


[imath]e^{i\pi+ln3}[/imath] beggs to differ. [/nitpick]
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