2^(x+3)-3^(x^2+2*x-6)=3^(x^2+2*x-5)-2^x
Кто знает, как решать?

2^(x+3)-3^(x^2+2*x-6)=3^(x^2+2*x-5)-2^x8*2^x + 2^x = 3^(x^2+2*x-6) + 3 * 3^(x^2+2*x-6)9 * 2^x = 4 * 3^(x^2+2*x-6)2^(x-2) = 3^(x^2+2*x-8)2 = 3^log(3,2), поэтому3^(log(3,2)*(x-2)) = 3^(x^2+2*x-8)log(3,2)*(x-2) = x^2+2*x-8x^2 + x*(2 - log(3,2)) - 8 + 2*log(3,2)=0D = (2 - log(3,2))^2 - 4 * (-8 + 2*log(3,2)) = 4 - 4*log(3,2) + (log(3,2))^2 + 32 - 8*log(3,2) = (log(3,2))^2 - 12*log(3,2) +36 = (log(3,2) - 6)^2x(1,2) = (log(3,2) - 2 +- (log(3,2) - 6)) / 2x1 = (log(3,2) - 2 - log(3,2) + 6) / 2 = 2x2 = (log(3,2) - 2 + log(3,2) - 6) / 2 = log(3,2) - 4