Question 5387
{{{1/(log(x))}}} <= {{{1/(log(sqrt(x+2)))}}} -- cross multiply. Not sure if this changes the inequality round though!


{{{(log(sqrt(x+2)))}}} <= {{{(log(x))}}} -- raise everything to power 2.


{{{sqrt(x+2)}}} <= {{{x}}} -- now square both sides


{{{x+2}}} <= {{{x^2}}}


{{{x^2 - x - 2}}} >= 0


(x+1)(x-2) >= 0


So, this is asking where is the quadratic greater or equal than zero ie greater or equal than the x-axis (y=0). Well, we know the roots are now x=-1 and x=+2, so the answer is x<=-1 or x>=2: assuming that the inequality doesn't swap round when we cross multiplies the logs in the first step.


Is that clear :-)?


jon.