X^2/(X+1)=a*e^x
Solve this algebric equation for different values of constant a (postiive, negative and 0)
Detailed explanation: YES
1
Expert's answer
2011-12-22T08:35:16-0500
Necessary condition x≠-1 a=0: so we have x^2/(x+1) = 0 => x^2 = 0 => x=0 a>0: take log of both parts of equation => => log(x^2) – log (x+1) = log a + log (e^x) => (log (e^x) = x) => log a = log (x^2) – log (x+1) – x a<0: we couldn’t solve with logarithms (a couldn’t be less than 0), so x^2/(x+1) ≈ x+(1/(x+1)) – 1 => => a = x/(e^x) + (1/e^x(x+1)) – 1/e^x
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