To solve this inequality, we can let y = ((1)/(7))^x.

The inequality becomes: y^2 - 8y + 1 <= 0

This is a quadratic inequality that can be solved by factoring or using the quadratic formula. Factoring, we get:

(y - 1)(y - 7) <= 0

This inequality holds true when y is between 1 and 7, inclusive.

Substitute back in for y:

((1)/(7))^x is between 1 and 7, inclusive.

We can then solve the inequality:

1 <= ((1)/(7))^x <= 7

Taking the natural logarithm of both sides:

ln(1) <= ln(((1)/(7))^x) <= ln(7)

0 <= xln(1/7) <= ln(7)

0 >= xln(1/7) >= ln(7)

0 >= x(-ln(7)) >= ln(7)

0 >= -xln(7) >= ln(7)

0 <= x <= ln(7)/(-ln(7))

0 <= x <= -1

Therefore, the inequality ((1)/(7))^(2x-1) - 8((1)/(7))^x + 1 <= 0 is true when x is between 0 and -1, inclusive.