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

For the discussion of math. Duh.

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

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

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)
"Now I shall have less distraction"
--Leonhard Euler, on losing sight in his right eye

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

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!)

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

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

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

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

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

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

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

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.

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

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.

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

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.)

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

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

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

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

Except for e-3, of course.

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

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

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

[imath]e^{i\pi+ln3}[/imath] beggs to differ. [/nitpick]
I edit my posts a lot and sometimes the words wrong order words appear in sentences get messed up.