When 2^x $$ \ge $$ 1

5 + 2^x –1 = 2^x (2^x – 2)

Let 2x = t

$$ \Rightarrow $$5 + t – 1 = t (t – 2)

$$ \Rightarrow $$ t = 4, – 1(rejected)

$$ \Rightarrow $$ 2x = 4

$$ \Rightarrow $$ x = 2

Now when 2^x < 1

5 + 1 – 2^x = 2^x (2^x – 2)

Let 2^x = t

$$ \Rightarrow $$ 5 + 1 – t = t (t – 2)

$$ \Rightarrow $$ 0 = t² – t – 6

$$ \Rightarrow $$ 0 = (t – 3) (t – 2)

$$ \Rightarrow $$ t = 3, – 2

2^x = 3, 2^x = – 2 (rejected)

$$ \therefore $$ Only one real roots.