An Exponential Equation | (x-3)^6=(3-x)^x

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[seanfraser]

I like this one. It looks really difficult if you just go in with brute-force algebra, but if you stop and think about it it’s actually very easy.

[yoav]

Nice! Now i understand why wolfram gives only x=2, 3 .👌💯

[raystinger]

Assume y=x-3. We get:
y^6=(-y)^(y+3)

y^3=-(-y)^y
-y^3=(-y)^y
(-y)^3=(-y)^y

And now the bases are equal, so we can solve the rest as seen on the video.

[vinda]

Right off the bat I got 2 as an answer from the video cover. 3 also works.

[angshukNag]

fourth roots of unity might yield complex solutions as well

[devondevon]

let n = x-3 then
-n = 3-x
and x =n+3 ;

hence
n^6 = (-n)^n+3
n^5 =(-1)^n+3
n^5 =-1
n=-1
hence -1 = x-3
x = 2 is a solution
from the original equation x also =3 since
(3-3)^6 = (3-3)^3 or 0=0
answer x =2 and x =3

[carlyet]

on inspection x== 6 is a solution and consider x<>3 1 = (-1)^x . (3 - x)^[x-6]

[Frank-kxhc]

Let X= 3-x then the equation is equivalent to : X^6=X^x

If X= 0 ===> x=3: x=3 is solution,
If X<>0 we can divide both sides by X^x, we obtain :
X^p=1, where p=6-x <====>
{• p=0 and X<>0 : x= 6 is solution,
• or X=1 and p exists: x=2 is solution,
• or X=-1 and p EVEN: x=4 is solution bcse p=6-4=2 is EVEN }
Thus S={3, 6, 2, 4}

This RULE is in:

[mcwulf]

Can also show x cannot be greater than 6 by way of contradiction. If x> 6 then we can factor:
(x-3)^6 ((x-3)^(x-6) - 1) = 0
The first bracket can't be 0 for x>6 and if the second is zero then
(x-3)^(x-6) = 1
Again this is not possible for x > 6.

[alextang]

This question is not difficult. For the first glance, set two situations first, one is x=3 and the other is x not=3. After that, the answer will be easily found. 😊😊😊😊😊😊

[mega_mango]

Случай нуля, любой другой случай, и случай единицы:
0^х = -0^у;
х = 3
k^х = -k^2n;
х =(?) 2n;
(2n-3)^⁶ =(?) -1^2n * (2n - 3)^2n
2n = x = 6;
±1^х = -1^2n; х = 4, 2

[rachit]

Before watching, one solution should be 4

[MichaelJamesActually]

Is it not mathematically defined to raise (-1) ^ 1 power?

[MichaelRothwell]

LHS=(x-3)⁶=(3-x)⁶, so our equation is equivalent to
(3-x)⁶=(3-x)ˣ.

First note that x=3 is a solution, since for x=3, LHS=0=RHS.

From now on we assume x≠3, so x-3≠0, and (3-x)ˣ≠0.

So (3-x)⁶=(3-x)ˣ
⇔(3-x)⁶÷(3-x)ˣ=1
⇔(3-x)⁶⁻ˣ=1

Hence either:
6-x=0
or 3-x=1
or 3-x=-1 and 6-x is such that (-1)⁶⁻ˣ>0
(Proof given below)

Hence x=2, x=4 or x=6.

Adding the first solution found, x=2, x=3, x=4 or x=6.

Proof of the three possibilities given above for (3-x)⁶⁻ˣ=1.

x≠3, since for x=3, LHS=0³=0.
(3-x)⁶⁻ˣ=1
⇔|(3-x)⁶⁻ˣ|=1 and (3-x)⁶⁻ˣ>0 (the 2nd condition is implicit in the following lines until it reappears)
⇔|3-x|⁶⁻ˣ=1 (using the result that if x and y are real and xʸ is defined as a real number then |xʸ|=|x|ʸ)
As x≠3, |3-x|>0, so we can take ln of each side
So |3-x|⁶⁻ˣ=1
⇔ln[|3-x|⁶⁻ˣ]=ln 1
⇔(6-x)ln|3-x|=0
⇔6-x=0 or ln|3-x|=0
⇔6-x=0 or |3-x|=1
⇔6-x=0 or (3-x=±1 and (3-x)⁶⁻ˣ>0)
⇔6-x=0 or 3-x=1 or (3-x=-1 and (3-x)⁶⁻ˣ>0)
⇔6-x=0 or 3-x=1 or (3-x=-1 and (-1)⁶⁻ˣ>0)

[AllanPoeLover]

Why can't the second equation X be bigger than 3 ?

[cientista]

For university students, the word that should be used is function and not equation.