0

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 $y =t^{\sqrt {2/\lg t}}$.

Clement C.
  • 67,323
ssss
  • 101
  • 1
    I can see a transformation, but I don't see much of a simplification. – MasB Mar 19 '17 at 21:35
  • @sss Please, don't delete your question once you get an answer -- one of the the goals of this website is to keep these questions and answers archived, so that they may benefit other people as well. – Clement C. Mar 28 '17 at 13:34

1 Answers1

0

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 $\lg t = \sqrt{(\lg t)^2}$.

Clement C.
  • 67,323
  • is there any way to go the other way :) ? @ClementC – ssss Mar 19 '17 at 21:36
  • @ssss Sure, do the steps in the other direction. $$2^{\sqrt{ 2 \lg t}} = 2^{\sqrt{\frac{2}{\lg t}\cdot (\lg t)^2}} = 2^{\sqrt{\frac{2}{\lg t}\cdot (\lg t)^2}} = 2^{\sqrt{\frac{2}{\lg t}}\sqrt{ (\lg t)^2}} = 2^{\sqrt{\frac{2}{\lg t}}\lg t} = t^{\sqrt{\frac{2}{\lg t}}}$$ But it's (a bit) harder to figure out the steps in that direction, which is why I suggested to start the other way around. – Clement C. Mar 19 '17 at 21:39
  • sorry if I'm being pedantic, but is "$\lg$" a common abbreviation? - never seen it used before – mrnovice Mar 19 '17 at 21:41
  • @mrnovice It is often/sometimes used for $\log_2$. – Clement C. Mar 19 '17 at 21:41
  • Ah okay, thanks for that it makes sense now – mrnovice Mar 19 '17 at 21:42
  • @ClementC. thanks but $2^{\sqrt{{{(2/lgt}) * (lgt)^2}}}$ is confusing and not sure how you got it – ssss Mar 19 '17 at 21:54
  • @ssss You have $2\lg t = \frac{2(\lg t)^2}{\lg t}$, and thus $\sqrt{2\lg t} = \sqrt{\frac{2(\lg t)^2}{\lg t}}=\sqrt{\frac{2}{\lg t}}\sqrt{(\lg t)^2} = \sqrt{\frac{2}{\lg t}}\lg t$. (We used $\lg t >0$ in the last step to write $\sqrt{(\lg t)^2}=\lg t$.) – Clement C. Mar 19 '17 at 21:56