An algebra problem by Lucas Nascimento

Algebra Level 2

Find the number of possible real values of x x satisfying 4 x − 5 ⋅ 2 x + 4 = 0. 4^x - 5 \cdot 2^x + 4 = 0 .


The answer is 2.

Lucas Nascimento by Lucas Nascimento , 20, Brazil

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1 solution

Alex G
Oct 21, 2016

Let u = 2 x u=2^x . The equation is then

u 2 − 5 u + 4 = 0 u^2-5u+4=0

Which can be factored into

( u − 1 ) ( u − 4 ) = 0 (u-1)(u-4)=0

Which has two solutions, at u = 1 u=1 and u = 4 u=4 . 2 \boxed{2}

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