Answer to Question #5580 in Algebra for Megha

Question

X^2/(X+1)=a*e^x
Solve this algebric equation for different values of constant a (postiive, negative and 0)
Detailed explanation: YES

Answer 1

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