Consider f(t) = ln t on the interval [1,x].
Notice that f is continous on [1,x] and differenciable on (1,x).
By Lagrange's Mean Value Theorem:
(f(x)-f(1))/(x-1) = f'(c) for c in (1,x)
Thus,
ln(x)/(x-1) = 1/c for c in (1,x)
Notice that
1/x< 1/c < 1/1
Thus,
1/x < ln (x) / (x-1) < 1
Multiply by denominator,
(x-1)/x < ln x < (x-1)
Consider,
f(t) = sqrt(1+t) on interval [24,x].
Thus, f is continous on [24,x] and differenciable on (24,x).
Use the Mean Value Theorem:
[sqrt(x+1) - sqrt(24+1)]/[x-24] = f'(c) = 1/2*sqrt(1+c) for c in (24,x).
Note that, 1/2*sqrt(1+c) < 1/2*sqrt(24+1) = 1/10
Thus,
[sqrt(x+1) - 5]/[x-24] < 1/10
Thus,
sqrt(1+x) < 5 + (x-24)/10