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Mathematics 56 Online
OpenStudy (anonymous):

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

ganeshie8 (ganeshie8):

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

ganeshie8 (ganeshie8):

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

ganeshie8 (ganeshie8):

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

ganeshie8 (ganeshie8):

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

OpenStudy (jiteshmeghwal9):

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

ganeshie8 (ganeshie8):

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

ganeshie8 (ganeshie8):

its a quadratic equation in t you can solve for t ?

OpenStudy (jiteshmeghwal9):

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