# What's wrong with my simplification of these exponentials?

The question states:

Simplify $$2^{\sqrt2}$$

I tried doing: $$2^{\sqrt2}=2^{(2^{\frac{1}{2}})}$$ $$=(2^2)^{\frac{1}{2}}$$ $$=4^{\frac{1}{2}}$$ $$=2$$ $$=2^1$$

Clearly this is ridiculous as I've just "proved" $$2^{\sqrt2}=2^1$$. Yet my proof seems quite convincing to me. Where did I go wrong? I tried debugging by trying a similar technique on $$2^4$$ and it worked fine.

$$2^{4}=2^{2^{2}}$$ $$=(2^2)^{2}$$ $$=4^{2}$$ $$=16$$ $$=2^4$$

• Your second equality is false. RHS is $2^{1}$ and LHS is $2^{\sqrt 2}$. Sep 28 at 9:42
• $a^{b^c}\neq(a^b)^c$. Sep 28 at 9:43
• People tend to overlook the fact that iterated exponentiation is not associative. As a matter of definition, $a^{b^c}$ means $a^{(b^c)}$.
– lulu
Sep 28 at 9:44
• Anyway, I think $2^{\sqrt2}$ is already as simple as it's going to get. I don't know what someone had in mind when they asked for it to be simplified. Sep 28 at 9:45
• The second example works because of $2^2=2\cdot 2=4$ Sep 28 at 10:59

Because $$x^{(y^z)}$$ (like you did in the first example) does not mean $$(x^y)^z$$. It works in your second example because by chance $$2^2=2\times 2$$.