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Math Help - Rational equation with X as an exponent, converting to a quadratic and solve w/log

  1. #1
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    Rational equation with X as an exponent, converting to a quadratic and solve w/log

    I'll just start out by showing the equation.
    y = (10^x + 10^-^x) / (10^x - 10^-^x)

    I had a similar problem. I'll show how I worked that one out so that you can see my train of thought on this one.

    y = (10^x + 10^-^x)/2
    2y = 10^x + 10^-^x

    Multiply both sides by 10^-^x

    2y10^x = 10^2^x +1

    Now it's a quadratic. Solving for 10^x, a=1, b=-2y, and c=1
    10^2^x - 2y^x +1 = 0

    I'll save both of us some time by not showing the quadratic equation steps.
    So, I now have:
    10^x = y +- sqrt(y^2-1)

    Now, I take the log of both sides:
    log10^x = log(y +- sqrt(y^2-1))

    So, x = log(y +- sqrt(y^2-1))

    Now, back to the problem at hand.

    I multiplied both sides of the original equation by 10^x - 10^-^x to get:
    y10^x - y10^-^x = 10^x - 10^-^x
    I then multiplied both sides by 10^-^x to get:
    y10^2^x - y = 10^2^x - 1

    I then bring everything to one side:
    y10^2^x - 10^2^x - y + 1 = 0

    But, I can't see how this is quadratic. Either I screwed up before this point (the most likely scenario), or I'm just not seeing the quadratic.

    Please help. Thanks.
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  2. #2
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    Quote Originally Posted by franzpulk View Post
    I then bring everything to one side:
    y10^2^x - 10^2^x - y + 1 = 0
    Now factor out 10^{2x}

    10^{2x}(y-1)- y + 1 = 0

    10^{2x}(y-1)= y - 1

    10^{2x}= \frac{y - 1}{y-1}

    10^{2x}= 1 \implies x=0
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  3. #3
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    That seems a bit easy for problems from my instructor. But, it looks good to me. Thanks for the help.
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  4. #4
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    Quote Originally Posted by pickslides View Post
    Now factor out 10^{2x}

    10^{2x}(y-1)- y + 1 = 0

    10^{2x}(y-1)= y - 1

    10^{2x}= \frac{y - 1}{y-1}

    10^{2x}= 1 \implies x=0
    0 isn't in the domain though? Since it would be division by 0?


    If we sub in x=0 into y = \dfrac{10^x + 10^-^x}{10^x - 10^{-x}} we get

    \dfrac{1+1}{1-1} = \dfrac{2}{0}
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  5. #5
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    x can equal zero in the domain as it's only an exponent. Anything raised to the power of zero is 1. 1 is allowed in the domain of the denominator, so x=0 is allowed.
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  6. #6
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    Quote Originally Posted by franzpulk View Post
    x can equal zero in the domain as it's only an exponent. Anything raised to the power of zero is 1. 1 is allowed in the domain of the denominator, so x=0 is allowed.
    You're doing 1-1 in the denominator since exponents come before subtraction

    Yours and Pickslides methods do work, but since 0 is not in the domain this equation has no solutions

    edit: Your denominator is 10^x - 10^{-x}. So for x =0 we get 10^0 - 10^{-0} and since 10^0 = 10^{-0} = 1 then the denominator is 1-1 = 0
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  7. #7
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    Good point. So, there's something else to this problem then.

    The example I showed on my first post was done in class. He said to make sure we paid attention to it. Then, he gives us homework, and while similar, it's nothing we've done before.

    I'm still thinking that I need to make it a quadratic and solve it that way. Otherwise, why would he waste his time in class showing it to us?

    Any tips? Also, don't rely on my math. Make sure you go by the original formula.

    Thanks.
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  8. #8
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    if the goal is to solve for x ...

    y(10^x - 10^{-x}) = 10^x + 10^{-x}<br />

    y(10^{2x} - 1) = 10^{2x} + 1

    10^{2x} \cdot y - y = 10^{2x} + 1

    10^{2x} \cdot y - 10^{2x} = y + 1

    10^{2x}(y - 1) = y+1

    10^{2x} = \frac{y+1}{y-1}

    2x = \log\left(\frac{y+1}{y-1}\right)

    x = \log{\sqrt{\frac{y+1}{y-1}}}
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  9. #9
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    Quote Originally Posted by franzpulk View Post
    I multiplied both sides of the original equation by 10^x - 10^{-x} to get:
    y10^x - y10^-^x = 10^x - 10^-^x
    I then multiplied both sides by 10^x to get:
    y10^2^x - y = 10^2^x - 1
    You seem to have had a sign change on the right hand side! When multiplying through by the denominator it looks like you changed your +10^{-x} on the right to -10^{-x}

    y10^x - y10^{-x} = 10^x + 10^{-x}

    You carried forward the error but used the correct working (and some examiners will award you the method marks for this)

    y10^{2x} - y = 10^{2x} + 1

    y10^{2x} - 10^{2x} = y + 1

    10^{2x}(y-1) = y+1

    10^{2x} = \dfrac{y+1}{y-1}

    2x = \log \left(\dfrac{y+1}{y-1}\right)

    x = \dfrac{1}{2}\log \left(\dfrac{y+1}{y-1}\right)

    Still no quadratic though
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  10. #10
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    @ e^(i*pi) - That sign change was a typo on this post only. I have that correct on my hardcopy. I appreciate the help. What was getting me was pulling the 10^2^x out. That and not using the quadratic. Thanks again for the help.

    @ skeeter - I'm assuming your last step was a typo? It should look like e^(i*pi)'s last step, correct? Not the square root like you showed? Thanks.
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  11. #11
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    Quote Originally Posted by franzpulk View Post
    @ skeeter - I'm assuming your last step was a typo? It should look like e^(i*pi)'s last step, correct? Not the square root like you showed?
    no typo ... review your log properties.
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  12. #12
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    @skeeter - I'm not finding anything regarding dividing a log and roots. You and e^(i*pi) have done two different things with the leading "2" coefficient:


    x = \dfrac{1}{2}\log \left(\dfrac{y+1}{y-1}\right)

    and,


    x = \log{\sqrt{\frac{y+1}{y-1}}}

    I'm still not sure which is the correct method.
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  13. #13
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    Quote Originally Posted by franzpulk View Post
    I'll just start out by showing the equation.
    y = (10^x + 10^-^x) / (10^x - 10^-^x)
    Let u= 10^x
    Then the equation becomes
    y= \frac{u+ \frac{1}{u}}{u- \frac{1}{u}}
    Multiplying numerator and denominator on the right by u
    y= \frac{u^2+ 1}{u^2- 1} so that (u^2- 1)y= u^2y - u^2= u^2+ 1

    (y- 2)u^2= 1
    u^2= \frac{1}{y- 2}
    u= 10^x= \frac{1}{\sqrt{y- 2}}
    x ln(10)= -ln(\sqrt{y- 2})
    Last edited by HallsofIvy; November 29th 2010 at 11:39 AM.
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  14. #14
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    Quote Originally Posted by franzpulk View Post
    @skeeter - I'm not finding anything regarding dividing a log and roots. You and e^(i*pi) have done two different things with the leading "2" coefficient:


    x = \dfrac{1}{2}\log \left(\dfrac{y+1}{y-1}\right)

    and,


    x = \log{\sqrt{\frac{y+1}{y-1}}}

    I'm still not sure which is the correct method.
    both are correct ...

    the "power" property of logarithms states that if y = a\log{x} , then y = \log{x^a}. The converse of this statement is also true.
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