The statements will be proved one at a time.
b>1
If b is greater than 1, the f(x)=logbx is over its entire .
In the graph, it can be seen that
f(x1)≥f(x2) x1≥x2. Considering the definition of
f(x), the statement is proven.
logbx1≥logbx2⇔x1≥x2
b<1
If b is greater than 0 and less than 1, then f(x)=logbx is over its entire domain.
In the graph, it can be seen that
f(x1)≥f(x2) if and only if
x1≤x2. Considering the definition of
f(x), the statement is proven.
logbx1≥logbx2⇔x1≤x2