If you take the derivative, you get e^(x^2 + 2x)*[2*(x+1)^2 + 1]. This is clearly greater than 0 for all x, therefore, the original function is strictly increasing i.e. there's at most 1 real solution. From there, you can pretty easily see that x = 0 works. If you want to motivate x = 0, note that 1/(1+x) converges to its geometric series for x belonging to (-1,1). So ln(1+x) can be described as the integral of that series on the same interval. Plugging in x = -1 and 1 into (x+1)e^(x^2+2x) yields 0 and 2e^3. So by IVT and the fact that the function is strictly increasing, there is a single point between x = -1 and x = 1 s.t. (x+1)e^(x^2+2x) = 1 i.e. we can treat ln(1+x) as the integral of the geometric series since our solution lies in the radius of convergence. So first take natural log of both sides ==> ln(x+1) + x^2 + 2x = 0 ==> ln(x+1) = -x^2 - 2x ==> x(1 - x/2 + x^2/3 - ... + ) = -x(x+2). The fact that both sides are a multiple of x, suggests that x = 0 is a potential solution. From there, you can check that is indeed the solution.
Another method without using Lambert W functions: e^(x+1)^2 = e^(x^2+2x +1)-->(x+1) = u and u*e^(u^2 -1) = 1 -->u*e^u^2 = e , multiplying both sides by u--->u^2*e^(u^2) = e*u therefore u=0 -->x=0 or x=-2 , replacing can be seen that the only solution is x=0.
Gotta love it when one can solve the Lambert function by eye without calculations.
If you take the derivative, you get e^(x^2 + 2x)*[2*(x+1)^2 + 1]. This is clearly greater than 0 for all x, therefore, the original function is strictly increasing i.e. there's at most 1 real solution. From there, you can pretty easily see that x = 0 works.
If you want to motivate x = 0, note that 1/(1+x) converges to its geometric series for x belonging to (-1,1). So ln(1+x) can be described as the integral of that series on the same interval. Plugging in x = -1 and 1 into (x+1)e^(x^2+2x) yields 0 and 2e^3. So by IVT and the fact that the function is strictly increasing, there is a single point between x = -1 and x = 1 s.t. (x+1)e^(x^2+2x) = 1 i.e. we can treat ln(1+x) as the integral of the geometric series since our solution lies in the radius of convergence.
So first take natural log of both sides ==> ln(x+1) + x^2 + 2x = 0 ==> ln(x+1) = -x^2 - 2x ==> x(1 - x/2 + x^2/3 - ... + ) = -x(x+2). The fact that both sides are a multiple of x, suggests that x = 0 is a potential solution. From there, you can check that is indeed the solution.
That's a really clever approach to finding the solution!
Another method without using Lambert W functions: e^(x+1)^2 = e^(x^2+2x +1)-->(x+1) = u and u*e^(u^2 -1) = 1 -->u*e^u^2 = e , multiplying both sides by u--->u^2*e^(u^2) = e*u therefore u=0 -->x=0 or x=-2 , replacing can be seen that the only solution is x=0.
Nice. And you don't really need to know about W to solve this.
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I also got x=0 as the only solution.
(x+1)e^((x+1)^2)=e.,.[x+1=t]...=te^(t^2)=e...[^2]...t^2e^(2t^2)=e^2...2t^2=W(2e^2)...t=√W(2e^2)/√2..=√2/√2=1...x=1-1=0
If W(te^t) = y and y >0, then there is only one solution :)
x = 0 or -2
Could you solve this
let f(x)=aˆx and g(x)=log_a(x)
Solve a so that these functions only touch once
(x+1)e^((x+1)^2) = e
2(x+1)^2 e^(2(x+1)^2) = 2e^2
2(x+1)^2 = 2
x + 1 = ±1
x = -2 (extraneous) or x = 0
Why are you squaring.
Why so many steps. Multiply both sides by e. Let y = x +1 =>. y*e^y^2=e. The only solution is y=1 or x = 0.
Cool!
x = 0 by inspection. I took the calculus route to show that the function was always increasing.
No. Bye.
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