Fractional Exponents (How is it done?)

AI Thread Summary
The discussion focuses on understanding how to express the exponent 2^(5/2) as a product of simpler terms, specifically 2^2 multiplied by 2^(1/2). Participants clarify that 2^(5/2) can be split into 2^(4/2) + 2^(1/2), which simplifies to 2^2 and the square root of 2, respectively. There is confusion regarding the notation and the proper way to handle fractional exponents, with some participants suggesting that the expression could be misinterpreted if not formatted correctly. The conversation emphasizes the importance of recognizing how fractional exponents relate to roots and integer powers. Overall, the thread provides insights into the manipulation of fractional exponents and their representation.
Bored
Messages
3
Reaction score
0
How does 2^5/2 become 2^2 multiplied by 2^1/2?
(The '^' means 'to the power of' so 2 to the power of 5/2. I am not sure how to write this as an exponent as this is my first post.)

2^5/2 = 2^2 × 2^1/2

So 2^2 = 4 and 2^1/2 means Square Root so there is a radical sign, so it becomes √2.

I tried to reverse engineer the solution but I'm still not sure how 2^5/2 makes it. I know you first Square root it so √2 then you put to the power of 5 so √2^5. This goes into decimals so I am confused on what to do next. I looked at the solution and I can see that both of them make 2^5/2 I just don't know how it was factorized into those specific numbers, because 4 x √2 = 4√2 and somehow that is √2^5? I know 2^5 also makes 32.

So I think I might of done something wrong or I don't know the correct method, can someone please attempt to help me and thank you for your time!
 
Mathematics news on Phys.org
Do you agree that x^(a+b)=(x^a) (x^b) for natural x,a,b?

edit, forgot a +
 
Last edited:
In my case would that be 2^5/2 (5 is a and 2 is b) = (2^5) (5^2)? I'm quite sure I'm doing something wrong so I think I can't agree on that even though I think its correct.

So what is a and b? If a and b are 5 and 2 they make (2^5) (5^2) making 800. I am kind of lost... Am I getting it wrong?

Another go at it I see that 2^5 = 32 and 5^2 = 25 - So do I put in the fraction making it 2^5/1 and 5^1/2?

Oh now I see it and the '+', Thank you for changing it
 
Last edited:
Do you agree that
2^4 = 2^2 * 2^2
3^{10}=3^4 * 3^6
 
Last edited:
Yes, Because when you are multiplying powers you add them together, but since I started fractional exponents I am very confused - probably because of the fractions.

So if you added 2^2 and 2^1/2 that would make 2^3/2 if i am correct or 2.5?

Okay so there's 2/1 + 1/2 = 2 5/2
So adding fractions you double it to make it 4/2 then add it to make it 5/2 making it 2^5/2.
So we done this backwards but how did 2^5/2 split up to make those 2?

Oh wait (Edit), So you could also spilt it up to something else? But it has to be a Fractional Exponents because so it can be combined? Not sure why it has to be a fractional exponent but I know you split the expression into the product of an integer exponent and the factor with a fractional coefficient less than one. I think the reason why this is done is so you can keep the root in the equation, otherwise it no longer has a fractional exponent.

Thank you for your valuable time, which you were very helpful!

PS. If I made any mistakes please correct (:
 
Last edited:
2+\frac{1}{2}=\frac{4}{2}+\frac{1}{2}=\frac{5}{2}
 
2^5/2 is problematic, it could mean ##\frac{2^5}{2}##. You can write it as 2^(5/2) or better [noparse]25/2[/noparse], which gets parsed as 25/2. And you can use LaTeX here.

You can split up the 5/2 like this:
$$\frac{5}{2}=\frac{4+1}{2}=\frac{4}{2}+\frac{1}{2}=2+\frac{1}{2}$$
And then use the rule for exponents written above.
 

Similar threads

Replies
10
Views
2K
Replies
2
Views
2K
Replies
5
Views
2K
Replies
2
Views
2K
Replies
41
Views
5K
Replies
5
Views
1K
Replies
59
Views
2K
Replies
20
Views
2K
Back
Top