what is 2^(4-1)^2 ?

(2 raised to (4-1) raised to 2)

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- Jan 8th 2009, 09:49 AMvscidexponent confusion
what is 2^(4-1)^2 ?

(2 raised to (4-1) raised to 2) - Jan 8th 2009, 10:04 AMmasters
- Jan 8th 2009, 10:11 AMvscid
- Jan 8th 2009, 10:18 AMmasters
- Jan 8th 2009, 10:23 AMvscid
- Jan 8th 2009, 05:12 PMHallsofIvy
$\displaystyle (2^{4-1})^2= (2^3)^2= 8^2= 64$ which is $\displaystyle 2^6$. That is true because $\displaystyle (a^b)^c= a^{bc}$

However, $\displaystyle 2^{(4-1)^2}$ which is how what you wrote should be interpreted, is equal to $\displaystyle 2^{3^2}= 2^9= 512$ - Jan 8th 2009, 07:01 PMvscid
- Jan 8th 2009, 07:30 PMmr fantastic
- Jan 9th 2009, 04:51 AMvscid
- Jan 9th 2009, 08:09 AMmasters
The rule is when you take a power to a power you multiply the exponents:

$\displaystyle (x^m)^n=x^{mn}$

Yours was a special case. Not only did you want to take the base to a power, you also wanted to take the power to a power.

$\displaystyle {2^3}^2$

In this case you must work from the top down. If you were putting this in a calculator you would have to enter it this way:

2^(3^2) = 512

The parentheses tell the calculator to perform this operation first to get 9, then take 2 to the 9th power to get 512.

If you entered it 2^3^2, you would get 64 because the calculator performs the operations from left to right: 2^3 is 8 and 8^2 is 64.

It's a little tricky when you stack exponents. Just remember to start at the top and work your way down.

Good luck. - Jan 9th 2009, 09:49 AMvscid
Above you have written: 2^(3^2) = 512.

I do not have the parenthesis in red in my original post. In fact, I should say I have no parenthesis (besides the inevitable (4-1) )

So it is actually 2^3^2.

My question was, if there are no parenthesis as above, then the value will be 2^9 right?