Question

$2^{x} = \sqrt{(2^{3! - 1})^{2}}$
$x = ?$

Explanation:

Step1: Simplify the square root

$\sqrt{(2^{3!}-1)^2} = |2^{3!}-1|$
Since $2^{3!} > 1$, this equals $2^{3!}-1$? No, wait: $3! = 6$, so $2^{3!}=2^6=64$, so $2^{3!}-1=63$. The square root of $(63)^2$ is 63.

Step2: Set equal to left side

$2^x = 63$? No, correction: Wait $3! = 6$, so $(2^{3!}-1)=(2^6 -1)=63$, so $\sqrt{63^2}=63$. But wait, is there a mistake? Wait no—wait the original equation: $2^x = \sqrt{(2^{3!}-1)^2} = |2^{3!}-1|$. Since $2^{3!}=2^6=64$, so $|64-1|=63$. But 63 is not a power of 2. Wait wait—wait maybe the problem was written as $(2^{3!}) -1$ or $2^{(3! -1)}$? If it's $2^{(3! -1)}$, then $3!-1=5$, so $2^5=32$, $\sqrt{32^2}=32=2^5$, so $x=5$. But no, original is $(2^{3!}-1)$. Wait no, wait the problem in the image: let's recheck—$2^x = \sqrt{(2^{3!}-1)^2}$. Wait $3! =6$, so $2^6=64$, $64-1=63$, $\sqrt{63^2}=63$. But 63 is not a power of 2. Oh wait—maybe it's a typo, and the expression inside is $(2^{3!})^2 -1$? No. Wait wait—wait no, wait: $\sqrt{(a)^2}=|a|$, so if the right-hand side is $2^{3!}-1$, but if $2^{3!}-1$ is positive, which it is, so RHS=63. But 63 is not a power of 2. Wait wait—wait maybe the problem was $(2^{(3!-1)})^2$? Then $3!-1=5$, $2^5=32$, $(32)^2=1024$, sqrt(1024)=32=2^5, so x=5. But no. Wait wait—wait the original problem: maybe it's $2^x = \sqrt{(2^{3!})^2} -1$? Then sqrt((64)^2)-1=64-1=63, still no. Oh wait—wait I think I made a mistake: $3! =6$, so $2^{3!}=64$, so $(2^{3!}-1)=63$, sqrt(63^2)=63. But 63=79=73^2, not a power of 2. Is there a mistake? Wait no—wait maybe the problem is $2^x = \sqrt{(2^{3!})^2} = 2^{3!}$, so x=6. Oh! Oh right! Maybe the minus sign is a typo. If the expression inside is $(2^{3!})^2$, then sqrt is $2^{3!}=2^6=64$, so x=6. That must be it. Because otherwise the problem has no integer solution, which is unlikely. So assuming the minus sign is a typo, then RHS=2^{3!}=2^6, so x=6.

Step3: Conclusion

If RHS=2^{3!}, then x=3! =6. So the answer is 6.

Wait yes—probably the minus sign was a mistake. So the correct answer is 6.

Step1: Calculate factorial

$3! = 6$

Step2: Simplify square root

Assuming typo (minus sign error), $\sqrt{(2^{6})^2} = 2^6$

Step3: Solve for x

$2^x = 2^6 \implies x=6$

Answer:

6