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1 ) I have this equation and don't know how $(1 + 2e^y)$ and $(1 + 2e^x)$ from each side of equation cancelled each other out and get the final answer $x = y$.

$$\frac{e^x}{1+2e^x}=\frac{e^y}{1+2e^y}$$

$$e^x(1 + 2e^y) = e^y(1 + 2e^x)$$

$$e^x = e^y$$

$$x = y$$

Question 2 - I don't know how the equation turned into fraction in the last step. Appreciate much for fast reply :)

$$y =\frac{e^x}{1+2e^x}$$

$$e^x = y + 2ye^x$$

$$e^x =\frac{y}{1-2y}$$ (How does $y + 2ye^x$ turned into this?)

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Check the concept of "one-to-one functions". If f(x)=f(y) leads to x=y then the function f is one-to-one. One way to check whether f is one -to-one is through its derivative. –  Maesumi Sep 28 '12 at 9:04

1 Answer 1

up vote 3 down vote accepted

If you need I can add in the missing steps. For question 1, you have

$e^x(1+2e^y)=e^y(1+2e^x)$

Clear the parentheses to obtain

$e^x+2e^{x+y}=e^y+2e^{x+y}$

Then just subtract $2e^{x+y}$. Now for problem 2, you have

$e^x=y+2ye^x$

$e^x-2ye^x=y$

$e^x(1-2y)=y$

Now just divide and you have your last line.

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Got it. Thank you so much! –  Michael Sep 28 '12 at 9:12

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