1. 125^2/5 2. -8^5/3 i have a third, but i cant find it...i will post later if i find it =]
S silverlining326 New member Joined Jan 25, 2006 Messages 16 Feb 19, 2006 #1 1. 125^2/5 2. -8^5/3 i have a third, but i cant find it...i will post later if i find it =]
T tristatefabricatorsinc Junior Member Joined Jan 31, 2006 Messages 60 Feb 19, 2006 #2 for one I believe it would be the fifth root of 125 to the 2nd power for the second question it would be the third root of 8 which is 2 to the fifth power
for one I believe it would be the fifth root of 125 to the 2nd power for the second question it would be the third root of 8 which is 2 to the fifth power
S silverlining326 New member Joined Jan 25, 2006 Messages 16 Feb 19, 2006 #3 but how do you do that? thats what i need the explaining on =]
T tristatefabricatorsinc Junior Member Joined Jan 31, 2006 Messages 60 Feb 19, 2006 #4 for instance... the 3rd root of 125 is 5, you take 5 * 5 * 5 and it = 125 the third root of 8 is 2 since 2 * 2 * 2 = 8 if you had 4^1/2 for example the 2 in the fraction means the square root of 4 and the top number means to the first power does this make sense?
for instance... the 3rd root of 125 is 5, you take 5 * 5 * 5 and it = 125 the third root of 8 is 2 since 2 * 2 * 2 = 8 if you had 4^1/2 for example the 2 in the fraction means the square root of 4 and the top number means to the first power does this make sense?
pka Elite Member Joined Jan 29, 2005 Messages 11,990 Feb 19, 2006 #5 \(\displaystyle \L \left( {125} \right)^{\frac{3}{5}} = \left( {5^3 } \right)^{\frac{3}{5}} = \left( {5^{\frac{6}{5}} } \right) = 5\left( {5^{\frac{1}{5}} } \right)\) \(\displaystyle \L \left( { - 8} \right)^{\frac{5}{3}} = \left( { - 2^3 } \right)^{\frac{5}{3}} = - 2^5 = - 32\)
\(\displaystyle \L \left( {125} \right)^{\frac{3}{5}} = \left( {5^3 } \right)^{\frac{3}{5}} = \left( {5^{\frac{6}{5}} } \right) = 5\left( {5^{\frac{1}{5}} } \right)\) \(\displaystyle \L \left( { - 8} \right)^{\frac{5}{3}} = \left( { - 2^3 } \right)^{\frac{5}{3}} = - 2^5 = - 32\)