A Nice Exponential Equation

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[icfj]

I think the left side tends to 1 when x tends to 1, and the right side is equal to 1, so x-->1 can be considered in certain way a solution (I know that x^x theds to 1 when x tends to 0).

[MSK_VLOG.]

If we take log on both sides, we get x =2

[deadded]

Bottom can be i and -i if top is a multiple of 4.
Check i
x= 1+i
x^2 -1 = 2i -1 does not work
Check -i
x=1-i
x^2-1 = -2i-1 does not work

[divitmaheshwari]

This can be easily be solved by logarithms. But idk why 2 is coming as an answer for him.

(x-1)^(x²-1)=1
x²-1= log 1 base (x-1)
x²-1=0
x=+-√1
Therefore, x=1, x=-1

[mathsquared]

1 can be written as (x-1)⁰ and now follow the powers and exponents rules to compare both sides and get x²-1=0 and then x=+- 1

[ulugbeknayimov]

Good luck with your job. Try make more video like this. I support you

[prosp]

A genius in the making. God bless you..I pray you find Christ Sewell❤️🥰

[mohamedhosam]

How to make this video please tell me what logical do you use

[당근-gr]

sometimes, for convenience, the 0 power of 0 is defined as 1

[ailtoncordeiro]

Cold be better than a "guess n' check"

[lukamedin]

If we ln both sides, do we get a quadric equation? :)

[carlinoiavarone]

1 is not possible because x - 1 must be bigger than 1, so two solution 2 and -1

[okabekun]

Bro why you dont allow 0^0
Its defined 😭

[jaimealday]

Excuse Sir, but your writing can barely seen.

[muisnotforyou]

x=1 should really be counted as a legitimate solution, since 0⁰=1 makes perfect sense from however you want to define exponentiation. I think most people fall victim to the misconception that limits would break this, although all it would mean is that some function is not continuous.

[CrsHOvrRiD]

Let me make an example:

Notice that the equation (x-1)^(x^2-1)=1 can be rewritten as this:
(x-1)^(x^2-1) = 1^1

Using the property that any number to the power of 1 is itself, we can simplify this to:

(x-1)^(x^2-1) = 1

Now, there are two cases to consider:
Case 1: (x-1)^(x^2-1) = 1 and x-1 = 1
If x-1 = 1, then x = 2.

Case 2: (x-1)^(x^2-1) = 1 and x-1 = -1
If x-1 = -1, then x = 0.

Therefore, the solutions to the equation is: (x-1)^(x^2-1)=1 are x = 0 and x = 2.