Be Careful With This Math | How To Find Two Roots To This Math Problem | Math Olympiad Preparation.

[@leolectrified5462]

Maybe an easier way to get X = 1 Since X^(X-1) = X^(1-X) By applying lin function ln(X^(X-1)) = ln(X^(1-X)) By using logarithmic properties ln(a^b) = ln(a)^b = b.ln(a) So, (X-1)ln(X) = (1-X)ln(X) Divide by ln(X) on both sides X-1 = 1-X So, 2X = 2, X = 1 (I wouldn't have got -1 using this method but I considered -1 since in this equation it has two solutions and usually if they're real solutions they're gonna be the positive and negative of a real number)

[@alucardthespy5539]

x^(x - 1) = x^(1 - x) Rewrite as... x^x / x = x / x^x Cross multiply x^2x = x^2 (meaning there are two answers +/-) 2x = 2 x = 1 x = [+/-]1

[@kalapalababurao1788]

Nice 👍

[@onlineMathsTV]

Thanks sir.

[@justabunga1]

Set the equation another way as x^(x-1)=x^(-(x-1)). Take the log on both sides as (x-1)ln(|x|)=-(x-1)ln(|x|). Adding -(x-1)ln(|x|) will bring us 2(x-1)ln(|x|)=0. This means that x=±1. Note that the absolute value here is to extend the possible solutions for this equation. Otherwise, if you try to graph those two equations, you will see that x=-1 is out of the domain because the function must be continuous and x is always going to positive, which is basically x>0.

[@ManojkantSamal]

Respected Sir, Good afternoon....

[@danielfranca1939]

I learnt something new honestly, thanks so much for this deep explanations sir. Much respect sir...🙏👏👏

[@AliHassan-hb1bn]

I did it in one minute and in my head.

[@AliHassan-hb1bn]

Exponential equation has one posotive solution and if you change into quadratic equation it has two real solutions.

[@AliHassan-hb1bn]

You can factorise or use root property , in both cases you reach two real solutions.

[@BartBuzz]

Also, x^(x-1)/x^(1-x) = 1 Then, x^(2x-2) = x^2•x^(-x) = x^2 = 1 So, x = ± 1

[@ElvisSaturn]

Sir, I think you make a mistake from 10:37 : (-1)-² = (-1)² 1/(-1)² = (-1)² 1/1 = 1 (Power applies for -1 with minus sign) Solution: x^(x-1) = x^(1-x) -> x^(x-1) = x^ -(x-1) -> x^(x-1) = 1/(x^(x-1)) -> x^(x-1) * x^(x-1) = 1 -> x^2(x-1) = 1: Because number multiplied by 2 always results in even number: x = 1 or x = -1 .... ((-1)^even number=1)

[@onlineMathsTV]

No, sir. So long the whole expression is not in a bracket, the power cannot affect the minus. Thanks sir.

[@balogunibraheem5677]

@@onlineMathsTV but x takes the value of -1. it's not that we have -x^x-1 but x^x-1

[@AliHassan-hb1bn]

How I changed into a polynomial as x^2-1=0

[@SidneiMV]

case 1 x - 1 = 1 - x => *x = 1* case 2 x = 1 [ repeated solution ] case 3 x = -1 and even exponent -1 - 1 = -2 and 1 - (-1) = 2 => *x = -1* is also a solution case 4 x = 0 and positive exponent 0 - 1 < 0 so x = 0 is not a solution

[@onlineMathsTV]

Well detailed sir.

[@balogunibraheem5677]

thanks for the time sir but I think x has the value of -1 for the case 2 and the initial equation is x^x-1 which we can say is the same as (x)^x-1 so I think the power should affect the minus

[@onlineMathsTV]

Here, if there is no bracket covering the minus, then the power can not affect the minus sir. You can do a research on that sir.

[@magnusjonsson4540]

The two students are correct. x = -1 = (-1).

[@WorldwideBibleClass-qr9jk]

🎉🎉🎉

[@musicsubicandcebu1774]

(x-1) = a, -(1-x) = -a xᵃ = x⁻ᵃ xᵃ/ x⁻ᵃ = 1 x²ᵃ = 1 . . . x = 1

[@maburwanejohannes4647]

If you can solve k^2 - 1 = 0, then you should be able to solve this

[@adinandrawardhana]

Too long & too complicated explanation! x = 1. Case solved!!!