Simplifying $2^{\sqrt {2\lg t}}$ should yield $t^{\sqrt {2/\lg t}}$, but the following is what I am getting:
$$\begin{align} y &= 2^{\sqrt {2\lg t}}\\ y &= 2^{\sqrt {\lg t{^2}}}\\ y &= 2^{({{\lg t{^2}})}^{1/2}}\\ \lg y &= \lg{2^{({{\lg t{^2}})}^{1/2}}}\\ \lg y &= (\lg t{^2})^{1/2} \cdot \lg 2\\ \lg y &= (\lg t{^2})^{1/2} \end{align}$$
and then I'm not sure if I am heading towards a correct solution or not because after the last step I'm really confused. I'm not sure how I could get $y =t^{\sqrt {2/\lg t}}$.
It might be easier to start from the other end:
For $t>1$, $$ t^{\sqrt{\frac{2}{\lg t}}} = 2^{\sqrt{\frac{2}{\lg t}}\lg t} = 2^{\sqrt{\frac{2}{\lg t}}\sqrt{ (\lg t)^2}} = 2^{\sqrt{\frac{2}{\lg t}\cdot (\lg t)^2}} = 2^{\sqrt{ 2 \lg t}} $$ where we first used (1) that $a^b = 2^{b\lg a}$ for any $a>0$ and $b$, and (2) then $\lg t >0$ (as $t>1$) to say that $\lg t = \sqrt{(\lg t)^2}$.