Define ln x = $\int_1^x dt/t$ for x > 0.
Then by the definition of the integral and the continuity of 1/x, ln x = (x-1)/c where $1 \le c \le x$ if x > 1 and $1 \ge c \ge x$ if x < 1. The biggest (x-1)/c can be if x >1 is if c = 1, which gives x -1 < x.
If x < 1 the expresson (x-1)/c is negative; whereas between 0 and 1 x is positive, so log x < x in that case also.
Now you may think I'm playing games with this problem, but in fact that definition of ln x is given by Richard Courant in his calculus book. Here is why he does it that way (besides that it makes problems like this one easy to do):
While usually the log is defined as the inverse of the exponential, that is a clumsy approach. It leaves one with complicated and/or finicky proofs about the continuity and differentiability of the exponential.
However, if you define the log as ln x = $\int_1^x dt/t$ you gain many advantages. All the properties of the logs can be derived directly and with little trouble from this definition. By this definition the log function is automatically differentiable. Then when you define the exponential function as its inverse, all the properties you want for it fall out of the corresponding properties of the log, including, of course, its differentiability.