A Nonstandard Equation With Exponentials

[@spelunkerd]

Gotta love it when one can solve the Lambert function by eye without calculations.

[@Blingsss]

Exponential equations always feel like puzzles with infinite twists. This video made an otherwise tricky concept surprisingly engaging! It also reminded me how helpful platforms like SolutionInn can be for breaking down complex topics.

[@miraj2264]

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.

[@SyberMath]

That's a really clever approach to finding the solution!

[@cav1928]

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.

[@mcwulf25]

Nice. And you don't really need to know about W to solve this.

[@aekben7312]

👍

[@scottleung9587]

I also got x=0 as the only solution.

[@giuseppemalaguti435]

(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

[@KennethChile]

If W(te^t) = y and y >0, then there is only one solution :)

[@rakenzarnsworld2]

x = 0 or -2

[@c0ckyb4st4rd]

Could you solve this let f(x)=aˆx and g(x)=log_a(x) Solve a so that these functions only touch once

[@rob876]

(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

[@alesnecas7410]

Why are you squaring.

[@alesnecas7410]

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.

[@nickarrizza]

Cool!

[@robertlunderwood]

x = 0 by inspection. I took the calculus route to show that the function was always increasing.

[@supeskrim]

No. Bye.

[@SyberMath]

😄