How do you solve 8^x <= 25?

Answer 1

consider 8^x = 25
take logs of both sides and use your log rules
xlog8 = log 25
x = log25/log8 = appr 1.55

so x ≤ 1.55

test some values:
x = 0, 8^0 = 1 which is ≤ 25
x = -5 , 8^-5 = .0000305.. which is ≤ 25
we can easily see that all negative values of x will work

let's try some numbers ≥ 1.55
e.g. let x = 2, 8^2 = 64 which is ≥ 25 , but we excluded x=2

so x ≤ appr 1.55