Question

By using an appropriate substitution , solve 2(16 to the power of x) - 5 (4 to the power of x ) + 2 = 0

Answer 1

\(\huge 2. 16^x - 5 . 4^x + 2 = 0\)

Answer 2

\(\huge 2 . (2^4)^x - 5 . 2^{2x} + 2 = 0\)

Answer 3

\(\huge 2 . (2^{2x})^2 - 5 . (2^{2x}) + 2 = 0\)

Answer 4

say

\(\huge 2^{2x} = t\)

Answer 5

do wht @ganeshie8 sir is doing u will probably gt ur answer:)

Answer 6

=>

\(\huge 2t^2 - 5t + 2 = 0\)

Answer 7

its a quadratic equation in t

you can solve for t ?