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Math Help - Solving exponential equations.

  1. #1
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    Solving exponential equations.

    Can someone please show me how to solve this?

    e^x - e^-x = 1
    2

    Also my professor told me to multiply each side by e^x.. can someone help? Thanks.

    so I've tried it and solved it and got x=2.41 and x= -.41 (which when I plug it in, it doesn't equal 1).

    e^X (e^x - e^-x) = 2e^x
    which then results to (Let A = e^X)

    A^2 - 2A -1 = 0.
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  2. #2
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    Quote Originally Posted by EliteNewbz View Post
    Can someone please show me how to solve this?

    e^x - e^-x = 1
    2

    Also my professor told me to multiply each side by e^x.. can someone help? Thanks.

    so I've tried it and solved it and got A=2.41 and A= -.41 (which when I plug it in, it doesn't equal 1).

    e^X (e^x - e^-x) = 2e^x
    which then results to (Let A = e^X)

    A^2 - 2A -1 = 0.
    Since e^x>0 for all x\in \mathbb{R} the variable A must be greater than zero. So the you can't use the negative solution.

    Solve for x: e^x=1+\sqrt{2}~\implies~x\approx 0.88137...
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  3. #3
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    Quote Originally Posted by EliteNewbz View Post
    Can someone please show me how to solve this?

    e^x - e^-x = 1
    2

    Also my professor told me to multiply each side by e^x.. can someone help? Thanks.

    so I've tried it and solved it and got x=2.41 and x= -.41 (which when I plug it in, it doesn't equal 1).

    e^X (e^x - e^-x) = 2e^x
    which then results to (Let A = e^X)

    A^2 - 2A -1 = 0.
    Solve the quadratic for A. As b^2-4ac = 8 there are two real solutions in A

    A = \frac{2+2\sqrt{2}}{2} = 1+\sqrt{2}
    A = 1-\sqrt{2}

    e^x = 1+\sqrt{2}  \: \rightarrow \: x = ln(1+\sqrt2)

    e^x = 1-\sqrt{2} \: \rightarrow x = ln(1-\sqrt2) = ln(\sqrt{2}-1)+i\pi
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  4. #4
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    Thank you! I must have done something wrong then!
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