# How do I simplify the following equation?

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• March 20th 2012, 10:30 PM
Paulo1913
How do I simplify the following equation?
http://latex.codecogs.com/gif.latex?...=-log(x+e^x)+c, 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 http://latex.codecogs.com/gif.latex?...{\sqrt{x+e^x}} but I dont understand how they got to this point, especially the square root part....
• March 20th 2012, 11:34 PM
earboth
Re: How do I simplify the following equation?
Quote:

Originally Posted by Paulo1913
http://latex.codecogs.com/gif.latex?...=-log(x+e^x)+c, 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 http://latex.codecogs.com/gif.latex?...{\sqrt{x+e^x}} 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:

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

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

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

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

$\displaystyle{y=\frac1{\sqrt{x+e^x}} \cdot c}$
• March 20th 2012, 11:37 PM
princeps
Re: How do I simplify the following equation?
Quote:

Originally Posted by Paulo1913
http://latex.codecogs.com/gif.latex?...=-log(x+e^x)+c, 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 http://latex.codecogs.com/gif.latex?...{\sqrt{x+e^x}} but I dont understand how they got to this point, especially the square root part....

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

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

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

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

$y= \frac{C}{\sqrt{x+e^x}}$
• March 21st 2012, 04:55 PM
HallsofIvy
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: $y= e^{log_e((x+e^x)^{-1/2}}e^{-c/2}= (x+ e^x)^{-1/2}e^{-c/2}$.