Range of $\frac{\log(x)}{x^{b}}$ This is from Berkeley problems. Actual question is this


  For which positive numbers $a,b,~ a &gt;1$ does the equation $\log_{a}(x)=x^{b}$ have positive solution for $x$? 


For solving this we can rewrite the expression as $\log a= \frac{\log x}{x^b}$. Hence the problem reduces to this for what values of $x$  $\log a $ lies in the range of $f(x)= \frac{\log x}{x^b}$ the solution says range of  $f(x)= (\infty,\frac{1}{be})$. Using geogebra I can verify this. But how to find range of such functions? All I know is range of $\log x$ in $(-\infty,\infty)$ and range of $\frac{1}{x^b}$ in set of all non zero reals.
To find the range of $f(x) = \frac{\log(x)}{x^b}$, it suffices to use the usual calculus techniques to confirm that: 


$y = f(x)$ has a vertical asymptote at $x = 0$
there exists an $r$ such that $f(x)$ is increasing over $0 &lt; x &lt; r$, and that $f(x)$ is decreasing over $x &gt; r$
Differentiate the fuction .... then find its maxima..... in this case the function is increasing for x less than $e^{1/b}$. The corresponding y for this x is ${1/be}$. Hence this the max. Value of the function. Hence the range is ${-infinity}$ to ${1/be}$.