Question 222910
y = {e^x} / {1 + 2e^x} 
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Cross multiply to get:
y + 2ye^x = e^x
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Get the e^x terms together:
2ye^x-e^x = -y
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Factor out the e^x:
e^x(2y-1) = -y
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Divide thru by 2y-1 to get
e^x = -y/(2y-1)
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Take the natural log to solve for x
x = ln[-y/(2y-1)]
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Cheers,
Stan H.