Question 65918
Given that 2^(x+2) + 2^x = 160

==> (2^x)*(2^2) + 2^x = 160  [ as x^(a + b) = x^a * x^b]

==> 2^x (2^2 + 1) = 160  [removing 2^x as common factor]

==> 2^x (4 + 1) = 160

==> 2^x(5) = 160

==> 2^x = 160/5  [dividing by 5 throughout]
==> 2^x = 32
==> 2^x = 2^5   [as 32 = 2^5]

as the bases are the same we equate the exponents.

==> x = 5


Good Luck!!!