2 to the power of root 2

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apple2357

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Does 2 to the power of root 2 mean anything?
It doesn't make any sense to me when thinking about basic indices?
 
Because its irrational? And it was to the power of a rational number a/b , it makes sense to me , like 8^(2/3) = cube root and then square? But what about root 2? whats the equivalent interpretation?
 
apple2357 said:
Because its irrational? And it was to the power of a rational number a/b , it makes sense to me , like 8^(2/3) = cube root and then square? But what about root 2? whats the equivalent interpretation?
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You're right that rational powers can be explained in terms of the integer powers and roots that you initially learn about (x^3 = x*x*x; x^-3 = 1/x^3; x^(1/2) = square root of x; and so on).

When you come to irrational powers, you need to move into calculus: we define 2^sqrt(2) as the limit of 2^x as x approaches sqrt(2). That is, as you calculate 2^1, 2^1.4, 2^1.41, 2^1.414, and so on, you approach a specific number, which we take to be 2^sqrt(2). You just successively approximate the value.

You can also do this using logarithms.
 
Dr.Peterson said:
You're right that rational powers can be explained in terms of the integer powers and roots that you initially learn about (x^3 = x*x*x; x^-3 = 1/x^3; x^(1/2) = square root of x; and so on).

When you come to irrational powers, you need to move into calculus: we define 2^sqrt(2) as the limit of 2^x as x approaches sqrt(2). That is, as you calculate 2^1, 2^1.4, 2^1.41, 2^1.414, and so on, you approach a specific number, which we take to be 2^sqrt(2). You just successively approximate the value.

You can also do this using logarithms.
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Thank you, that's very helpful. So we are now looking at things in a different way!
 
apple2357 said:
Thank you, that's very helpful. So we are now looking at things in a different way!
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Right. Once you get into irrational numbers, you can extend many concepts like exponents using an assumption of continuity, which makes them much more powerful.
 
Dr.Peterson said:
Right. Once you get into irrational numbers, you can extend many concepts like exponents using an assumption of continuity, which makes them much more powerful.
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Thanks again. Is this the same thinking behind logs with negative bases or is that a simple No No?
 
apple2357 said:
Thanks again. Is this the same thinking behind logs with negative bases or is that a simple No No?
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Logs with negative bases are too messed up to be used at all. Consider that a negative base raised to an even power will be positive; to an odd power will be negative; to a fractional power like 1/2 will be non-real; to a fractional power like 1/3 will be negative; and so on. As a result, you can't even graph y = (-2)^x, so trying to find what x has to be to get a given value for y is just impossible. Nothing is continuous, or even continuously defined!
 
Dr.Peterson said:
You're right that rational powers can be explained in terms of the integer powers and roots that you initially learn about (x^3 = x*x*x; x^-3 = 1/x^3; x^(1/2) = square root of x; and so on).

When you come to irrational powers, you need to move into calculus: we define 2^sqrt(2) as the limit of 2^x as x approaches sqrt(2). That is, as you calculate 2^1, 2^1.4, 2^1.41, 2^1.414, and so on, you approach a specific number, which we take to be 2^sqrt(2). You just successively approximate the value.

You can also do this using logarithms.
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Hi, how can we calculate 2 to the power of root 2 by using logarithms? Isn’t it the Gelfond - Schneider constant?
 
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Farzin said:
Hi, how can we calculate 2 to the power of root 2 by using logarithms? Isn’t it the Gelfond - Schneider constant?
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How does its having a name (and having been proved to be irrational) affect the question? We can't calculate it exactly, any more than we can calculate [MATH]\sqrt{2}[/MATH] exactly; but I don't think that was the question -- it was whether it is a number at all, as I read the OP. And clearly it is.

A calculator in effect uses logarithms to find a value for it.
 
Dr.Peterson said:
How does its having a name (and having been proved to be irrational) affect the question? We can't calculate it exactly, any more than we can calculate [MATH]\sqrt{2}[/MATH] exactly; but I don't think that was the question -- it was whether it is a number at all, as I read the OP. And clearly it is.

A calculator in effect uses logarithms to find a value for it.
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Yes, I know that wasn’t the question and because I didn’t want to open a new thread, I just asked if it is possible to use logarithm to find approximate value of it. Thank you
 
Here is one way: [MATH]2^\sqrt{2} = \left(e^{ln(2)}\right)^\sqrt{2} = e^{ln(2)\sqrt{2}}[/MATH]. Now evaluate that.

Or, of course, just use a calculator, which does at least something like that internally: [MATH]2^\sqrt{2} = 2.6651441426902251886502972498731...[/MATH]
 
Dr.Peterson said:
Here is one way: [MATH]2^\sqrt{2} = \left(e^{ln(2)}\right)^\sqrt{2} = e^{ln(2)\sqrt{2}}[/MATH]. Now evaluate that.

Or, of course, just use a calculator, which does at least something like that internally: [MATH]2^\sqrt{2} = 2.6651441426902251886502972498731...[/MATH]
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Thank you.
 
apple2357 said:
Does 2 to the power of root 2 mean anything?
It doesn't make any sense to me when thinking about basic indices?
Click to expand...
I've written this before but can't find it now so I will repeat.
I presume that you have been introduced to the power notation as \(\displaystyle x^2= x\cdot x\), \(\displaystyle x^3= x\cdot x\cdot x\), etc. with "\(\displaystyle x^n\) mean "x multiplied by itself n times."

Obviously that requires that n be a positive integer. It just doesn't make sense to "multiply x by itself -1 times" or (your case) "multiply x by itself \(\displaystyle \sqrt{2}\) times" or even "multiply x by itself 0 times".

But there are many times we would like to use them. And, hey, this is mathematics- everything depends on definitions. We are free to define things as we want. As long as everyone agrees we are free to define "\(\displaystyle x^{\sqrt{2}}\)" as we like.

Now, \(\displaystyle x^n\), with n a positive integer, has two nice properties. One is that \(\displaystyle (x^n)(x^m)= x^{n+m}\) because \(\displaystyle x^n\) has n copies of x multiplied, \(\displaystyle x^m\) has m copies, so \(\displaystyle (x^n)(x^m)\) has n+ m copies of x multiplied. The other is that \(\displaystyle (x^m)^n= x^{mn}\). To see that, imagine \(\displaystyle x^m\) as a row of m x's. Taking that to the m power, we are multiplying m of those rows- think of a rectangle with m rows, each with n x's so there are mn x's.

As I said, we are free to define "\(\displaystyle x^r\)" as we please but it would be nice to do it so that those rules, \(\displaystyle (x^n)(x^m)= x^{n+ m}\), and \(\displaystyle (x^m)^n= x^{mn}\) were still true.

"0" is of course the "additive identity", n+ 0= n for any n. In particular \(\displaystyle x^nx^0=x^{n+ 0}= x^n\) so we want \(\displaystyle x^0= 1\) and that is how we define \(\displaystyle x^0\) for any x.
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Every positive integer has a "negative" such that \(\displaystyle n+ (-n)= 0\) so we can write \(\displaystyle x^{n}x^{-n}= x^{n- n}= x^0= 1\) so we want \(\displaystyle x^{-n}\) to be the "multiplicative inverse" of \(\displaystyle x^n\) which is \(\displaystyle \frac{1}{x^n}\).​
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That takes care of the integers. The next set of numbers is the rational numbers. A rational number is a number of the form \(\displaystyle \frac{m}{n}\) where m and n are integers. Now we want to define x to a fractional power so that \(\displaystyle (x^m)^n= x^{m+ n}\). In particular, we want \(\displaystyle (x^{\frac{1}{n}})^n= x^{\frac{n}{n}}= x^1= x\). That means we want \(\displaystyle x^{1/n} = \sqrt[n]{x} \), the nth root of x. Then, of course, \(\displaystyle x^{\frac{m}{n}}= \left(x^{\frac{1}{n}}\right)^m= \left( \sqrt[n]{x} \right)^m= \sqrt[n]{x^m} \)​
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The irrational numbers, like your \(\displaystyle \sqrt{2}\), cannot be defined "algebraically" as above- they have to be defined "analytically" with some kind of limiting process. The simplest is that every irrational number can be defined as the limit of a sequence of rational numbers. That is what we mean when we say that every irrational number iis For example, "\(\displaystyle \pi\)" is "3.141596..."- it is the limit of the sequence of rational numbers, 3, 3.1, 3.14, 3.141, 3.11415, etc. \(\displaystyle \sqrt{2}= 1.4142...\). That is, \(\displaystyle \sqrt{2}\) is the limit of the sequence of rational numbers 1, 1.4, 1.41, 1.414, 1.4142.​
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And we define x to an irrational power to make \(\displaystyle x^y \), for variable y, a continuous function. We define \(\displaystyle x^{\sqrt{2}}\) as the limit of the sequence \(\displaystyle x^1\), \(\displaystyle x^{1.4}\), \(\displaystyle x^{1.41}\), etc.​
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Now, you can't actually calculate the limit of that sequence (and neither can your calculator) but you can approximate it by taking an approximation to \(\displaystyle \sqrt{2}\). My calculator says that \(\displaystyle \sqrt{2}\) is equal to 1.4142135623730950488016887242097. It isn't, of course, but that is the rational number approximation my calculator is using. And \(\displaystyle 2^\sqrt{2}\) is given as 2.6651441426902251886502972498731. Again, that is not the actual value of \(\displaystyle 2^\sqrt{2}\) but is the best approximation it can give.​
 
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