# How do you find the relative extrema for f(x) = x - log_4x?

Apr 26, 2016

The minimum of $f \left(x\right)$ is$f \left(\frac{1}{\ln 4}\right) = 0.957$, nearly.

#### Explanation:

Use ${\log}_{b} a = {\log}_{c} a {\log}_{a} c$

$\ln x = {\log}_{e} x = {\log}_{4} x {\log}_{e} 4 = {\log}_{4} x \ln 4$.

${\log}_{4} x = \ln \frac{x}{\ln} 4$.

Now, $f \left(x\right) = x - \ln \frac{x}{\ln} 4$

$f ' = 1 - \frac{1}{x \ln 4} = 0$, when $x = \frac{1}{\ln} 4$

$f ' ' = \frac{1}{{x}^{2} \ln 4} > 0$, for all x, as $\ln 4 = 1.3863 > 0$..

Thus,$f \left(\frac{1}{\ln} 4\right)$ is the minimum of f(x).

$f \left(\frac{1}{\ln} 4\right) = \frac{1}{\ln} 4 - \ln \frac{\frac{1}{\ln} 4}{\ln} 4 = \frac{1 + \ln \left(\ln 4\right)}{\ln} 4 = 0.957$, nearly.-