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Math Help - How do I simplify the following equation?

  1. #1
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    How do I simplify the following equation?

    , to be in the form y=? where c is the constanf of the integration I have just done (not needed for the simplification)

    The answer is apparently but I dont understand how they got to this point, especially the square root part....
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  2. #2
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    Re: How do I simplify the following equation?

    Quote Originally Posted by Paulo1913 View Post
    , to be in the form y=? where c is the constanf of the integration I have just done (not needed for the simplification)

    The answer is apparently but I dont understand how they got to this point, especially the square root part....
    1. I'm going to use the usual notation for \log_e(x)=\ln(x)

    2. Your equation becomes:

    2\ln(y)=-\ln(x+e^x)+c

    \ln(y)=-\frac12 \cdot \ln(x+e^x)+c

    \ln(y)=\ln\left(\frac1{\sqrt{x+e^x}} \right)+c

    \displaystyle{e^{\ln(y)}=e^{\ln\left(\frac1{\sqrt{  x+e^x}} \right)+c}}

    \displaystyle{y=\frac1{\sqrt{x+e^x}} \cdot c}
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  3. #3
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    Re: How do I simplify the following equation?

    Quote Originally Posted by Paulo1913 View Post
    , to be in the form y=? where c is the constanf of the integration I have just done (not needed for the simplification)

    The answer is apparently but I dont understand how they got to this point, especially the square root part....
    2\ln y=-\ln (x+e^x) + C

    \ln y= \frac{-1}{2} \cdot \ln(x+e^x) + C

    \ln y = \ln \frac{1}{\sqrt{x+e^x}}+\ln C

    \ln y = \ln  \frac{C}{\sqrt{x+e^x}}

    y= \frac{C}{\sqrt{x+e^x}}
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  4. #4
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    Re: How do I simplify the following equation?

    Generally speaking, to solve the equation f(y)= A for y, you apply inverse function of y to both sides: f^{-1}(f(x))= f^{-1}(A). The definition of "inverse function" is that f^{-1}(f(x))= x and f(f^{-1}(x))= x so that gives y= f^{-1}(A).

    And one of the fundamental things you should have learned about exponentials and logarithms is that the logarithm, base a, is the inverse function to the exponential function, [tex]a^x[tex].

    From 2log_e(y)= -log_e(x+ e^x)+ c, the first thing you should do is divide by 2 to get log_e(y)= -(1/2)log_e(x+e^x)- c/2 and then y= e^{-(1/2)log_e(x+e^x)+ c} and then use the "laws of exponentials" and "laws of logarithms: y= e^{log_e((x+e^x)^{-1/2}}e^{-c/2}= (x+ e^x)^{-1/2}e^{-c/2}.
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