Решить с одз и и заменой
log^5 (7x-1)=log^5 (10x-13)
log2^2 (2-x)+5log^2 (2-x)=6

log^5 (7x-1)=log^5 (10x-13)

одз

7x - 1 > 0  x>1/7

10x - 13 > 0  x> 13/10

x∈(13/10, +∞)

7x - 1 = 10x - 13

3x = 12

x = 4

log2^2 (2-x)+5log^2 (2-x)=6

одз  2-х > 0   x < 2

log^2(2 - x) = t

t^2 + 5t - 6 = 0

t1 = 1

log^2(2 - x) = 1

2 = 2 - x

x = 0

t2 = -6

log^2(2 - x) = -6

2 - x = 2^-6

x = 2 - 1/2^6 = 1 63/64