# Consider the function $f(x)$ $=$ $\log_m(x) + \log_x(m)$ defined for $x>1$ , find the minimum value of $f(x)$ by applying the AM-GM Inequality.

Question: Consider the function $f(x)$ $=$ $\log_m(x) + \log_x(m)$ defined for $x>1$

For a fixed value of $m>1$, find the minimum value of $f(x)$ by applying the AM-GM Inequality.

What I have started:

$$f(x) = \log_m(x) + \log_x(m)$$

$$f(x) = \frac{\ln(x)}{\ln(m)} + \frac{\ln(m)}{\ln(x)}$$

$$f(x) = \frac{\ln^2(x)+\ln^2(m)}{\ln(x)\times\ln(m)}$$

Having trouble finding a way to apply the AM-GM Inequality? Also I cannot use calculus to solve this question.

Use AM-GM for $$f(x)=\frac{\ln (x)}{\ln (m)}+\frac{\ln(m)}{\ln(x)}$$
• @dydxx: Set $a=(\ln x)/(\ln m),b=(\ln m)/(\ln x)$ for $a+b\ge 2\sqrt{ab}$. Mar 19 '16 at 8:11