Expand the equation a bit to get
[tex]2^{2x} - 3\left(2^{x+2}\right) + 32 = 0[/tex]
[tex]2^{2x} - 3\left(2^x \cdot 2^2\right) + 32 = 0[/tex]
[tex]2^{2x} - 12 \left(2^x\right) + 32 = 0[/tex]
[tex]\left(2^x\right)^2 - 12 \left(2^x\right) + 32 = 0[/tex]
Substitute y = 2ˣ and you'll notice this is really a quadratic equation in disguise. You end up with
[tex]y^2 - 12y + 32 = 0[/tex]
which is easily solved by factorizing,
[tex](y - 8) (y - 4) = 0[/tex]
[tex]\implies y - 8 = 0 \text{ or }y - 4 = 0[/tex]
[tex]\implies y = 8 \text{ or }y = 4[/tex]
Now, 2³ = 8 and 2² = 4, so
• if y = 2ˣ = 8, then x = 3; otherwise,
• if y = 2ˣ = 4, then x = 2
