Question

Solve the inequality 4•4^x - 7•2^x ≤ 2

Answer 1

4•4^x - 7•2^x ≤ 2
4(2^x)^2 - 7•2^x ≤ 2

consider 4(2^x)^2 - 7(2^x) = 2
let 2^x = y , so we have
4y^2 - 7y - 2 = 0
(4y + 1)(y - 2) = 0
y = -1/4 OR y = 2

consider y = 2, then 2^x = 2 -----> x = 1
consider y = - 1/4 then 2^x = -1/4, which is not possible

so we only have x = 1 to consider, and remember we changed ≤ to =
So we have only to consider x > 1 and x < 1
we know x = 1 will work,

take a value of x < 1 , how about x = 0
LS = 4(4^0) - 7(2^0)
= 4 - 7 = -3, which ≤ 2

take a value of x > 1, let's pick x = 2
LS = 4(4^2) - 7(2^2)
= 64 - 28 , which > 2
so the only solution we have is x ≤ 1