Bonus Problem #2

Case 1: $2x^2 = 0$

$\Rightarrow x = 0$ but if $x = 0 \Rightarrow x^{2x^2} = 0^{2*0^2} \neq 1$

Case 2: $x = 1$

$\Rightarrow x^{2x^2} = 1^{2*1^2} = 1$

Case 3: $x = -1$ for even exponent

$\Rightarrow x^{2x^2} = (-1)^{2*(-1)^2} = 1$

So, they have two value.

Note $x^{2x^2}=(x^2)^{x^2}=1^1$ so $x^2=1 \Rightarrow x= \pm 1.$

Sravanth Chebrolu Proof that the cases are the few ones mentioned in the solution:

Let $0<a \not = 1$ be another answer (by false assumption).Then we'd have:

$a^{2a^2}=1 \rightarrow (2a^2) log \space {a} = log \space 1 \rightarrow (2a^2) log\space {a} = 0$

$\rightarrow 2a^2=0 \lor log \space{a}=0 \rightarrow a=0 \lor a=1$

Which is indeed a contradiction that proves that $\forall a \ge 0 \leftrightarrow a=0 \lor a=1$

Also for $x < -1$ we'll prove that the equation cannot hold:

$x < -1 \rightarrow x^2 > 1 \rightarrow (x^2)^{x^2} > 1^{x^2} = 1 \rightarrow x^{2x^2} > 1$

And hence we're done.

By the way,how are you feeling?Are thing starting to get better?

Well if ${ x }^{ 2x^{2} }=1\quad if\quad x=1,\quad { 1 }^{ 2 }=1$ Considering the exponent is two multiplied by a number, and the answer is one, we can assume that the exponent cannot be higher than one. However, we can still test for negative one. ${ x }^{ 2x^{ 2 } }=1,\quad if\quad x=(-1),\quad { x }^{ 2x^{ 2 } }={ -1 }^{ 2 }=1$

0^0 is not undefined. It is 1. The "correct" answer is two, with x being equal to either 0 or 1, but there could also be three solutions, with x being equal to -1.

Similarly we can just simplify it to

x^{2x^{2}} - 1 = 0

And just following through with that and expanding, we have

(x^{x^{2}} + 1) (x^{x^{2}} - 1)

Giving 2 solutions of 1, -1

(xᵡ² + 1)(xᵡ² ̶ 1) = 0 .... maybe that's enough... I hope.

$x^{2x^2} = (x^2) ^ {x^2} =1={1}^{1} \\$ so $\\ x^2=1 \\ x= 1,\ or\ x=-1$ $\\$ So, they have $\boxed{two}$ value

1 and -1 because (-1)^2(-1)^2=1 and also anything power 1 is 1 itself.

Note that if $|x|>1$ , $x^2>1$ and this way, $(x^2)^{x^2}>1$. So, $|x|=1$ because $x$ is a integer number.

Case 1: 1^2*1²=1

Case 2: -1^2*(-1)²=1

0^0 is undefined.

x^(2x^2)=1 or, log(x^(2x^2))=log1 or, (2x^2)logx=0 [as, log1=0] eighther, (2x^2)=0 or, logx=0 or, x=0 or x=1 that means x has two values

There are in fact 2 solutions:

-1,1 are all integers and solve the equation as one can easily verify.

Note: $0^0$ is often considered indeterminate.