how would i do this... :oops: i just blanked out! [2^(2a)2^(a+2)] / [2^(3a)]
T Tascja New member Joined Mar 4, 2006 Messages 46 Mar 4, 2006 #1 how would i do this... i just blanked out! [2^(2a)2^(a+2)] / [2^(3a)]
skeeter Elite Member Joined Dec 15, 2005 Messages 3,216 Mar 4, 2006 #2 when you multiply two numbers with the same base, you add the exponents when you divide two numbers with the same base, you subtract the exponents now ... you try this one. multiply the numerator first, then divide by the denominator.
when you multiply two numbers with the same base, you add the exponents when you divide two numbers with the same base, you subtract the exponents now ... you try this one. multiply the numerator first, then divide by the denominator.
skeeter Elite Member Joined Dec 15, 2005 Messages 3,216 Mar 4, 2006 #4 show what you did to get the value 1.
T Tascja New member Joined Mar 4, 2006 Messages 46 Mar 4, 2006 #5 [2^(2a)2^(a+2)] / [2^(3a)] = [2^(3a+2)] / [2^(3A)] = 1^2 =1 ???? is that right??
skeeter Elite Member Joined Dec 15, 2005 Messages 3,216 Mar 4, 2006 #6 Tascja said: [2^(2a)2^(a+2)] / [2^(3a)] = [2^(3a+2)] / [2^(3A)] this is good! = 1^2 this should 2^2, because 2^(3a+2)/2^(3a) = 2^[(3a+2)-(3a)] =1 ???? is that right?? Click to expand...
Tascja said: [2^(2a)2^(a+2)] / [2^(3a)] = [2^(3a+2)] / [2^(3A)] this is good! = 1^2 this should 2^2, because 2^(3a+2)/2^(3a) = 2^[(3a+2)-(3a)] =1 ???? is that right?? Click to expand...
T Tascja New member Joined Mar 4, 2006 Messages 46 Mar 4, 2006 #7 oo ok... so its supposed to be 4 well thank u sooo much AGAIN!!!
i just blanked out!