How do you solve $2^{x} + 2^{2 x} = 72$ $2^(x) + 2^(2x) = 72$ ?

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Answer 1 · 2016-06-22T08:51:21

$x = 3$ $x = 3$

Taking $y = 2^{x}$ $y = 2^x$ we have

$y + y^{2} = 72$ $y+y^2=72$

Solving for $y$ $y$ we get

${y = - 9}, {y = 8}$ ${y = -9},{y = 8}$

but $y = 2^{x}$ $y = 2^x$ must be positive then we follow with

$y = 2^{x} = 8 \to x = 3$ $y = 2^x = 8->x = 3$

How do you solve 2^(x) + 2^(2x) = 722x+22x=722^(x) + 2^(2x) = 72?