Решите уравнение 4^2x-3-4*2^2x-1+48=0

4 ^ (2 * x - 3) - 4 * 2 ^ (2 * x - 1) + 48 = 0; 4 ^ (2 * x) * 4 ^ (- 3) - 4 * 2 ^ (2 * x) * 2 ^ (- 1) + 48 = 0;4 ^ (2 * x)/4 ^ 3 - 4 * 2 ^ (2 * x)/2 + 48 = 0;4 ^ (2 * x)/4 ^ 3 - 2 * 2 ^ (2 * x) + 48 = 0;4 ^ (2 * x)/4 ^ 3 - 4 ^ x + 48 = 0;4 ^ (2 * x)/4 ^ 3 * 64 - 2 * 4 ^ x * 64 + 48 * 64 = 0; (4 ^ x) ^ 2 - 128 * 4 ^ x + 3072 = 0;4 ^ x = (128 - √4096)/(2·1) = (128 - 64)/2 = 64/2 = 32; 4 ^ x = (128 + √4096)/(2·1) = (128 + 64)/2 = 192/2 = 96; { 4 ^ x = 32; 4 ^ x = 96;{ 2 ^ (2 * x) = 2 ^ 5; x = log4 96; { x = 2.5; x = log4 96.