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Math Help - natural exponential function with two solutions

  1. #1
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    natural exponential function with two solutions

    e^(2x)-6e^x+5=0 has two solutions, what are they?

    I tried this:
    1. moved +5 to the other side so there wouldnt be a ln 0 for the next step
    2. applied ln to both sides [2x-6x=ln(-5)]
    3. combined like terms = -4x=ln(-5)
    4. isolated x = x=ln(-5)/-4

    obviously this is wrong

    the only thing i can think of that would yield two answers is the quadratic formula, but i don't know if thats the right route or how to set that up

    edit still stuck here. any tips on where to start?

    edit2: Ok. I just looked at this problem and realized that e^0 would work for one answer. I assume the other I can work out now.

    edit3: still stuck. tried everything i know
    Last edited by thedoge; January 25th 2007 at 08:22 PM.
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  2. #2
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    Quote Originally Posted by thedoge View Post
    e^(2x)-6e^x+5=0 has two solutions, what are they?
    ...
    Hello,

    to solve this equation you have to use substitution:

    e^{2x}-6e^x+5=0
    (e^{x})^2-6e^x+5=0

    Now set e^x=y. Your equation becomes now:

    (y)^2-6y+5=0 which have the solutions y = 1 or y = 5

    Now re-substitute:

    e^x=1 \Longrightarrow x = 0 or

    e^x=5 \Longrightarrow x = \ln(5)

    EB
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  3. #3
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    Quote Originally Posted by thedoge View Post
    e^(2x)-6e^x+5=0 has two solutions, what are they?

    I tried this:
    1. moved +5 to the other side so there wouldnt be a ln 0 for the next step
    2. applied ln to both sides [2x-6x=ln(-5)]
    Your problem was step 2.
    e^{2x} - 6e^x + 5=0

    e^{2x} - 6e^x = -5

    ln \left ( e^{2x} - 6e^x \right ) = ln(-5)

    Two points about this:
    1) ln(-5) doesn't exist. The domain for the ln function is  (0, \infty).

    2) ln \left ( e^{2x} - 6e^x \right ) \neq ln(2x - 6x). You are trying to mix a valid rule with a mistaken one here. Yes, ln \left ( e^{2x} \right ) = 2x, but ln(a^b + c) cannot be simplified in general.

    -Dan
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