J jsvlad New member Joined Sep 14, 2011 Messages 3 Sep 14, 2011 #1 simple exponents question. please see attached image. top: 2^(A+B) so 2a +2 on top 2a+2 - ab = 2a+2-a(a+2) .. = 2a+2-aa-2a = 2-a? ... a=1, b= 3? 2*2^3 = 32.. 2^4 / 2^6 = 1/2^2 = 1/4 must be 4.. q23 answer 4. i can't do this without plugging in numbers. i was hoping somebody can clarify. thanks..
simple exponents question. please see attached image. top: 2^(A+B) so 2a +2 on top 2a+2 - ab = 2a+2-a(a+2) .. = 2a+2-aa-2a = 2-a? ... a=1, b= 3? 2*2^3 = 32.. 2^4 / 2^6 = 1/2^2 = 1/4 must be 4.. q23 answer 4. i can't do this without plugging in numbers. i was hoping somebody can clarify. thanks..
J jsvlad New member Joined Sep 14, 2011 Messages 3 Sep 14, 2011 #2 Denis said: Will you please post something that can be understood? Click to expand... i apologize if it is unclear.. ill be more clear.. can somebody show me how to solve this problem without plugging in numbers!? thanks!
Denis said: Will you please post something that can be understood? Click to expand... i apologize if it is unclear.. ill be more clear.. can somebody show me how to solve this problem without plugging in numbers!? thanks!
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Sep 14, 2011 #3 jsvlad said: can somebody show me how to solve this problem without plugging in numbers!? Click to expand... We know that \(\displaystyle b=a+2\) \(\displaystyle \dfrac{2^a\cdot 2^b}{2^{ab}}=\dfrac{2^a\cdot 2^{a+2}}{2^{a(a+2)}}=\dfrac{2^{2a+2}}{2^{a^2+2a}}=\dfrac{2^2}{2^{a^2}}\)
jsvlad said: can somebody show me how to solve this problem without plugging in numbers!? Click to expand... We know that \(\displaystyle b=a+2\) \(\displaystyle \dfrac{2^a\cdot 2^b}{2^{ab}}=\dfrac{2^a\cdot 2^{a+2}}{2^{a(a+2)}}=\dfrac{2^{2a+2}}{2^{a^2+2a}}=\dfrac{2^2}{2^{a^2}}\)