Question

(2 la puterea 2017+2 la puterea 2016+2 la puterea 2015) : (2 la puterea 2010+2 la puterea 2009+2 la puterea 2008)

Answer 1

[$\frac{\left(\left(2^{2017+2}\right)^{2016+2}\right)^{2015}}{\left(\left(2^{2010+2}\right)^{2009+2}\right)^{2008}}=\frac{\left(\left(2^{2019}\right)^{2018}\right)^{2015}}{\left(\left(2^{2010+2}\right)^{2009+2}\right)^{2008}}=\frac{\left(\left(2^{2019}\right)^{2018}\right)^{2015}}{\left(\left(2^{2012}\right)^{2011}\right)^{2008}}=2^{85166074}$