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....

Re: How do I simplify the following equation?

Quote:

Originally Posted by

**Paulo1913**

1. I'm going to use the usual notation for $\displaystyle \log_e(x)=\ln(x)$

2. Your equation becomes:

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

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

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

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

$\displaystyle \displaystyle{y=\frac1{\sqrt{x+e^x}} \cdot c}$

Re: How do I simplify the following equation?

Quote:

Originally Posted by

**Paulo1913**

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

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

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

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

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

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: $\displaystyle f^{-1}(f(x))= f^{-1}(A)$. The **definition** of "inverse function" is that $\displaystyle f^{-1}(f(x))= x$ and $\displaystyle f(f^{-1}(x))= x$ so that gives $\displaystyle y= f^{-1}(A)$.

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