SOLUTION: I don't know how to type this, so I'll try to explain it. Square root symbol, with 6 outside (but inside the little section), and inside the square root is another square root sym

Algebra ->  Rational-functions -> SOLUTION: I don't know how to type this, so I'll try to explain it. Square root symbol, with 6 outside (but inside the little section), and inside the square root is another square root sym      Log On


   

Question 174247: I don't know how to type this, so I'll try to explain it.
Square root symbol, with 6 outside (but inside the little section), and inside the square root is another square root symbol. Inside this second square root symbol is a 2. Now subtract from that: square root with 12 inside the little part, and inside the square root is 2 with exponent of 13.
Hope this makes sense.
Thanks!
Found 2 solutions by ilana, jim_thompson5910:
Answer by ilana(307) About Me  (Show Source):
You can put this solution on YOUR website!
So I believe you would say this as "the sixth root of the square root of 2 minus the twelfth root of 2 to the thirteenth power". Are you supposed to estimate this using a calculator? If not, I think the most you can simplify it is 2^(1/12) - 2^(13/12). You can do that because, for instance, the square root of 2 is 2^(1/2). So "the sixth root of the square root of 2" is (the square root of 2)^(1/6), which is (2^(1/2))^(1/6). So you multiply the exponents and get 2^1/12. Then you work similarly with the second part.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Is the expression root%286%2Csqrt%282%29%29-root%2812%2C2%5E13%29 ???


root%286%2Csqrt%282%29%29-root%2812%2C2%5E13%29 Start with the given expression


root%286%2Csqrt%282%29%29-root%2812%2C2%5E%2812%2B1%29%29 Break up 2%5E%2813%29 to 2%5E%2812%2B1%29


root%286%2Csqrt%282%29%29-root%2812%2C2%5E%2812%29%2A2%5E%281%29%29 Break up the exponent (see note below)


root%286%2Csqrt%282%29%29-root%2812%2C2%5E%2812%29%29%2Aroot%2812%2C2%5E%281%29%29%29 Break up the square root (see note below).


root%286%2Csqrt%282%29%29-2%2Aroot%2812%2C2%29 Take the 12th root of 2%5E12 to get 2 (see note below).


root%286%2C2%5E%281%2F2%29%29-2%2Aroot%2812%2C2%29 Convert from radical notation to exponential notation. So sqrt%282%29=2%5E%281%2F2%29


%282%5E%281%2F2%29%29%5E%281%2F6%29-2%2Aroot%2812%2C2%29 Convert from radical notation to exponential notation. So root%286%2C2%5E%281%2F2%29%29=%282%5E%281%2F2%29%29%5E%281%2F6%29


%282%5E%281%2F2%29%29%5E%281%2F6%29-2%2A2%5E%281%2F12%29 Convert from radical notation to exponential notation. So root%2812%2C2%29=2%5E%281%2F12%29


2%5E%281%2F12%29-2%2A2%5E%281%2F12%29 Multiply the exponents 1%2F2 and 1%2F6 to get 1%2F12


Take note that the GCF is 2%5E%281%2F12%29


2%5E%281%2F12%29%281-2%29 Factor out the GCF 2%5E%281%2F12%29


2%5E%281%2F12%29%28-1%29 Subtract


root%2812%2C2%29%28-1%29 Convert from exponential notation to radical notation. So 2%5E%281%2F12%29=root%2812%2C2%29


-root%2812%2C2%29 Rearrange the terms.



So root%286%2Csqrt%282%29%29-root%2812%2C2%5E13%29=-root%2812%2C2%29 (Note: the right side reads the negative 12th root of 2)



Notes:

I used the following identities to simplify the expression:


1) x%5E%28y%2Bz%29=x%5Ey%2Ax%5Ez


2) sqrt%28x%2Ay%29=sqrt%28x%29%2Asqrt%28y%29


3) root%28n%2Cx%5En%29=x