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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.

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Use AM-GM for $$f(x)=\frac{\ln (x)}{\ln (m)}+\frac{\ln(m)}{\ln(x)}$$

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  • $\begingroup$ I'm still confused how to apply it :( $\endgroup$ Mar 19 '16 at 8:09
  • $\begingroup$ @dydxx: Set $a=(\ln x)/(\ln m),b=(\ln m)/(\ln x)$ for $a+b\ge 2\sqrt{ab}$. $\endgroup$
    – mathlove
    Mar 19 '16 at 8:11
  • $\begingroup$ Ah yes, thank you for that. I had trouble identifying a and b. Thank you so much :) $\endgroup$ Mar 19 '16 at 8:13

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