Question 59749
2^(x^2) = 32(2^4x)
==> 2^(x^2) = (2^5)(2^4x)  [32 = 2^5]
==> 2^(x^2) = 2^(5+4x)  [law of exponents: x^a * x^b = x^(a+b)]
AS the bases are the same on both the sides, the exponents can be equated.
==> x^2 = 5+4x
==> x^2-4x = 5  [subtracting 4x from both the sides]
==>x^2 - 4x - 5 = 0 [subtracting 5 from either side]
==>x^2 + x - 5x - 5= 0 [splitting the middle term]
==>x(x+1) - 5(x+1) = 0 [taking the common factor out]
==>(x + 1)(x - 5) = 0
==> x + 1 = 0    or x - 5 = 0
==> x = -1    or x = 5

Ans: x = -1, 5

Good Luck!!!