Here's one method: From \(\displaystyle \sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}= 2^x\), square both sides: \(\displaystyle 2\sqrt{2\sqrt{2\sqrt{2}}}= 2^{2x}\). Divide both sides by 2 so \(\displaystyle \sqrt{2\sqrt{2\sqrt{2}}}= 2^{2x- 1}\).
Now square again: \(\displaystyle \sqrt{2\sqrt{2}}= 2^{2(2x- 1)}= 2^{4x- 2}\).
Again, divide by 2 to get \(\displaystyle \sqrt{2\sqrt{2}}= 2^{4x- 3}\).
Square a third time: \(\displaystyle 2\sqrt{2}= 2^{8x- 6}\).
Again, divide by 2: \(\displaystyle \sqrt{2}= 2^{8x- 7}\)
Square a fourth time: \(\displaystyle 2= 2^1= 2^{16x- 14}\) so we must have 16x- 14= 1.
Another method. Use the facts that \(\displaystyle \sqrt{a}= a^{1/2}\), that \(\displaystyle (a^p)^q= a^{pq}\) and \(\displaystyle (a^p)(a^q)= a^{p+q}\).
So \(\displaystyle 2\sqrt{2}= 2^1(2^{1/2})= 2^{3/2}\)
\(\displaystyle \sqrt{2\sqrt{2}}= (2^{3/2})^{1/2}= 2^{3/4}\)
\(\displaystyle 2\sqrt{2\sqrt{2}}= 2^1(2^{3/4})= 2^{7/4}\)
\(\displaystyle \sqrt{2\sqrt{2\sqrt{2}}}= 2^{7/8}\)
\(\displaystyle 2\sqrt{2\sqrt{2\sqrt{2}}}= 2^{15/8}\)
Finally, \(\displaystyle \sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}= 2^{15/16}\).