#### marsh

##### New member
The solution is right there too but I just need to understand what is happening

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#### Subhotosh Khan

##### Super Moderator
Staff member
The solution is right there too but I just need to understand what is happening

Exactly which step you don't understand?

The first step was to re-write the inner-most √2 as 2½

and continue re-writing from there!

#### HallsofIvy

##### Elite Member
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, $$\displaystyle \sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}= 2^{15/16}$$.

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