Range of $\frac{\log(x)}{x^{b}}$

Question

This is from Berkeley problems. Actual question is this

For which positive numbers $a,b,~ a >1$ 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.

Answer 1

$$f(x) = x^{-b} \ln(x) $$ $$f'(x) = (-b)x^{-b-1} \ln(x) + x^{-b-1} =x^{-b-1} (1-b \ln(x))$$

So there will be a critical pt. where

$$ \ln(x)=\frac 1b \implies x=e^{1/b} $$

This will be the global max whenever $b>0$

The max value for $f(x)$ is

$$f( e^{1/b} ) = ( e^{1/b})^{-b} \ln(e^{1/b})=e^{-1}(1/b)=\frac 1{be} $$

Answer 2

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 $0 < x < r$, and that $f(x)$ is decreasing over $x > r$

Answer 3

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 ${1/be}$.