A Nice and Interesting Exponential Equation

[@satrajitghosh8162]

Case 1 : x^2 - x - 1 = 1 ( x -2) ( x + 1) = 0 : x = 2, -1 Case 2 : x^2 - x - 1 = - 1 and x ^2 - x = even x ( x - 1) = 0 Hereby x = -1, 0, 1, 2

[@timothybohdan7415]

You just proved that knowledge of math and the methods of Sybermath are more powerful than Wolfram Alpha.

[@thunderpokemon2456]

No in some situations w is god tier lvl

[@SyberMath]

They are! 😄

[@rdspam]

Why? Wolfram comes up with the same answers. And, fwiw, I came up with all 4, staring at it while in bed with my iPad, in about 3 minutes.

[@coolasitta3337]

You can also use logic to solve this: if a^b =1, it is only possible in 3 cases: 1) a=1 2) a=-1 and b=even number 3) b=0 Now you get 3 case scenarios and solve for each and you will get 4 real solutions

[@mystychief]

3) b=0 AND a NOT 0.

[@coolasitta3337]

@@mystychief Yes that's true. I forgot to mention it, my bad.

[@ProactiveYellow]

Only analytically. Just accept that 0⁰=1 and come to the abstract algebra side 😛

[@coolasitta3337]

@@ProactiveYellow 😂😂

[@SyberMath]

@@mystychief a can be 0 😁

[@innovbrain]

(t-1)^t=1, t*ln(t-1)=ln(1), t*lnt=0. Either t=0, means x^2-x=0, x=0 or, x=1 or, t=1, ×^2-x=1, either x=1 or x=0.

[@maxvangulik1988]

case 1: x^2-x=0 x(x-1)=0 x=0,x=1 case 2: x^2-x-1=1 x^2-x-2=0 x=(1+-3)/2 x=2,x=-1 case 3: x^2-x-1=-1 and x^2-x is an even integer x^2-x=0 and it's just case 1 again x=-1,0,1,2

[@yogesh193001]

You should set it =-1. It would've been fun. Euler would love you!

[@SyberMath]

@@yogesh193001 😁

[@Qermaq]

Hey, this is something that's puzzling me and maybe it would make a good problem. Maybe not. Consider the sequence 0, 1, 4, 15, 56, 209, 780, 2911, 10864, .... s.t. a(0) = 0, a(1) = 1, and when n < 1 a(n) = 4a(n-1) - (n-2). Is it possible to construct a function s.t. f(n) = a(n) but without reference to previous values? So f(1) = 1, f(2) = 4, f(3) = 15, etc. (BTW this sequence is all integers x where 3x^2 + 1 = k^2 (k is an integer). If you use (a(n), a(n-1)) as m and n in Euclid's Pythagorean triples parameterization, you get close approximations of the 30-60-90 right triangle where 2a + 1 = c.)

[@anotherelvis]

Google for: The On-Line Encyclopedia of Integer Sequences. Just type in your terms and get the answer

[@Ahwke]

Greatly🎉❤

[@Roq-stone]

Please, zero^zero has no defined value. This is much like assigning infinity a value. Infinity/infinity doesn’t equal one just like zero/zero doesn’t equal one.

[@SyberMath]

@@Roq-stone they are not the same thing. 0^0 is an empty product and it’s 1

[@scottleung9587]

Got 'em all!

[@SyberMath]

nice!

[@DarekKoczwara]

It took me 5 seconds to solve it. Since the right side is 1, then any base but 0 to power 0 is 1. Therefore x^2-x=0. x(×-1)=0, x=0 & x=1. No logs are needed. Just saying....

[@SidneiMV]

x² - x = u (u - 1)ᵘ = 1 u = 0 => x² - x = 0 => x(x - 1) = 0 *x = 0* ∨ *x = 1* u - 1 = 1 => u = 2 => x² - x - 2 = 0 x = (1 ± 3)/2 => *x = 2* ∨ *x = -1* u - 1 = -1 ∧ u is even u = 0 *x = {2, 1, 0, -1}*

[@ДмитрийЗайцев-ш9о]

The exponential function is defined only for a non-negative base, so x=0 and 1 are not solutions

[@jpolowin0]

That isn't true for integer exponents, even in just the world of non-complex numbers. (-½)² is well-defined, for example: (-½) × (-½). Dealing with non-integer exponents requires complex math, but still gives solutions. x² - ½x -1 = 0 is just as solvable, via the binomial equation, as 2x² - x -2 = 0, despite the "b²" term.

[@yuryp6975]

​​@@jpolowin0But normally the problem would state whether you are looking for real/integer/complex solutions, though. So if you solve for real numbers you would exclude negative base

[@jpolowin0]

@@yuryp6975 Nope. Raising negative numbers to integer powers is valid for any base in _any_ of the "usual" number sets (real, integer, complex, etc.). In integers, (-1)² is valid. In rationals, (-½)² is valid. In irrational numbers, (-π)² is valid (though of course its exact value can only be approximated). This is because raising things to integer powers is well-defined: _a^n_ = _a_ multiplied by itself _n_ times.

[@yuryp6975]

​@@jpolowin0Agree with regards to integer exponent. So if the problem asked for integer solutions only (diophantine equation), there wiyld be no issues. If solving for real numbers, seens the finction shoud be defined across the domain.

[@jpolowin0]

@@yuryp6975 There are only three cases when _xʸ_ can be equal to 1. (a) _y_ = 0, for any _x._ (b) _x_ = 1, for any _y._ (c) _x_ = -1, when _y_ is any even integer. The "even integer" restriction in case (c) is implicit; that's the way numbers work. I'm rusty enough in my math skills that I'm uncertain if other cases might be possible in the complex domain, but I don't think there are. Within the reals... that's just the way it is. The function requires complex math for non-integers, but not for integers. Regardless of how you feel it "should" be for reals, the function is defined for integer exponents even when the base is negative. Its value is the negative number _x_ multiplied by itself _y_ times.

[@chukwuemekaajima8373]

For (x^2-x-1)^(x^2-x) = 1 I would do: (x^2-x-1)^(x^2-x) = (x^2-x-1)^0 Now both sides have the same base Then: x^2 - 1 = 0; x(x-1) = 0; x 0 or x = 1; x = (0,1). Finding the solution(s) For x = 0; then (0^2-0-1)^(0^2-0)=1; 1=1; For x = 1; then (1^2-1-1)^(1^2-1)=1; 1=1; Therefore, x(0, 1) are solutions.

[@Ahwke]

❤❤❤

[@rakenzarnsworld2]

x^2-x=0 x = 0 or 1

[@GaganR-qe7iu]

JEE PYQ question

[@phill3986]

😎✌️👍✌️👍😎

[@GillesF31]

I did something else to get 2 and -1 ... (x² - x - 1)^(x² - 1) = 1 get k = x² - x if k = x² - x then (x² - x - 1)^(x² - 1) = 1 becomes: (k - 1)^k = 1 k - 1 = 1^(1/k) recall: 1^n = 1 k - 1 = 1 k = 2 k = 2 and k = (x² - x) => 2 = (x² - x) 2 = x² - x x² - x - 2 = 0 Δ = (-1)² - 4·1·(-2) = 1 + 8 = 9 √Δ = ±√9 = ±3 • root #1: x = (-(-1) + 3)/(2·1) = 4/2 = 2 • root #2: x = (-(-1) - 3)/(2·1) = -2/2 = -1 /// final results: ■ x = 2 ■ x = -1 🙂

[@GIFPES]

For Christ sake!!! Anything tp the 0 power is 1; so just make the exponent equals to 0.

[@AbuUbaydah320]

you should do this suppose that X=x^2-x you will have (X-1)^X=1=(2-1)^2 that X=2 and x^2-x=2 x^2-x-2=0 => x=-1 or x=2

[@prollysine]

case 1 , || x^2-x=0 & x^2-x-1 not zero , or x^2-x not zero & x^2-x-1=1 || , x^2-x=0 , x-1=0 , x=1 , --> 1^2-1-1= -1 , test , (-1)^0=1 , OK , case 2 , (x^2-x)*ln(x^2-x-1)=ln(1) , ln(1)=0 , x2-x=0 --> x=1 , ln(x^2-x-1)=0 --> if x^2-x-1=1 , x^2-x-2=0 , x= 2 , -1 , test x=2 , x^2-x-1=4-2-1 --> =1 , OK , x=-1 , x^2-x-1=(-1)^2-(-1)-1 --> =1 , OK , x=2 , x^2-x=4-2 --> =2 , not zero , OK , x= -1 , (-1)^2-(-1)=2 , not zero , OK , solu , x= 1 , 2 , -1 ,

[@dwm1943]

Well, Mr Syber, I have not even looked at your solution, yet, though it will doubtless be elegant and accurate. But there is some silly behaviour going on in your class... Now look, you lot. The expression in the bracket differs by 1 from the exponent, to we could write it t^(t+1)., and this will equal 1 only if t=1, or if t=-1 (with even power), or if the exponent = 0, which actually it will do if t=-1.. If t=1, then x=2 or x= -1 two solutions and if t=-1, then x=0 or x=1, two more solutions Four solutions, therefore, x = 2, or 1, or 0 or -1. Complex solutions? Who knows? De Moivre, perhaps. And if Lambert's getting in on this act, I'm out of here.

[@SyberMath]

😀

[@ДмитрийЗайцев-ш9о]

(-3)^3=-27, but (-3)^3=(-3)^(6/2)=729^(1/2)=27 consequently 27=-27

[@broytingaravsol]

easy