x=2     —>   and 

x=6   —>    and 



so the answer lies between 2 and 6…   plug in the remaining values of x (3,4,5) and you will find it.

—

Here there are only three additional values to check.  If however you ever find you have many more values to check (say you are using trial and error and you know "x=1" is too low while "x=100" is too high) then you can use a binary search:  x=1 too low, x=100 too high, x=50 would be your next guess. If that were too high you'd use x=25 next, if x=50 were too low, you'd use x=75 as your next guess, etc.   You will converge on the answer pretty quickly.   

—

Another approach to problems like this would be to take the log( ) of both sides of your equation and isolate the log()'s on one side of the equation (even if 'x' appears on both sides).  The reason is the side with the log() functions is likely to grow much more slowly than the other side of the equation so as you guess values, one side changes quickly, the other side stays close to the same value.



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