# Verify if the solution of the equation is real 4^x-2^x=12.

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The equation we have is 4^x-2^x=12

4^x - 2^x = 12

=> 2^2^x - 2^x = 12

let 2^x = y

=> y^2 - y = 12

=> y^2 - y - 12 = 0

=> y^2 - 4y + 3y - 12 = 0

=> y(y - 4) + 3(y - 4) = 0

=> (y + 3)(y - 4) = 0

y = -3 and y = 4

2^x = -3 and 2^x = 2^2

As 2^x cannot be negative we have only one root.

**The real solution of the equation is only x = 2**

We notice that 4^x = 2^2x

We'll re-write the equation:

2^2x - 2^x - 12 = 0

We'll replace 2^x by t:

t^2 - t - 12 = 0

We'll apply quadratic formula:

t1 = [1+sqrt(1 + 48)]/2

t1 = (1 + 7)/2

t1 = 4

t2 = -3

But 2^x = t1 => 2^x = 4 <=> 2^x = 2^2

Since the bases are matching,we'll apply one to one rule:

x = 2.

2^x = t2 => 2^x = -3 impossible!

**The equation has only one real solution, namely x = 2.**