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^x

8*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), поэтому

3^ (log (3,2) * (x-2)) = 3^ (x^2+2*x-8)

log (3,2) * (x-2) = x^2+2*x-8

x^2 + x * (2 - log (3,2)) - 8 + 2*log (3,2) = 0

D = (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) ^2

x (1,2) = (log (3,2) - 2 + - (log (3,2) - 6)) / 2

x1 = (log (3,2) - 2 - log (3,2) + 6) / 2 = 2

x2 = (log (3,2) - 2 + log (3,2) - 6) / 2 = log (3,2) - 4