# Answer to Question #5580 in Algebra for Megha

Question #5580

X^2/(X+1)=a*e^x

Solve this algebric equation for different values of constant a (postiive, negative and 0)

Detailed explanation: YES

Solve this algebric equation for different values of constant a (postiive, negative and 0)

Detailed explanation: YES

Expert's answer

Necessary condition x≠-1

a=0:

so we have x^2/(x+1) = 0 => x^2 = 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)

=>

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=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**Need a fast expert's response?

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