Browse Questions

# Prove that the logarithmic function is strictly increasing on $(0, \infty).$

Toolbox:
• A function $f(x)$ is said to be a strictly increasing function on $(a,b)$ if $x_1 < x_2\Rightarrow f(x_1) < f(x_2)$ for all $x_1,x_2\in (a,b)$
• If $x_1 < x_2\Rightarrow f(x_1) > f(x_2)$ for all $x_1,x_2\in (a,b)$ then $f(x)$ is said to be strictly decreasing on $(a,b)$
• A function $f(x)$ is said to be increasing on $[a,b]$ if it is increasing (decreasing) on $(a,b)$ and it is increasing (decreasing) at $x=a$ and $x=b$.
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly increasing on $(a,b)$ is that $f'(x) > 0$ for all $x\in (a,b)$
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly decreasing on $(a,b)$ is that $f'(x) < 0$ for all $x\in (a,b)$
Step 1:
Let $f(x)=\log x$
Differentiating w.r.t $x$ we get,
$f'(x)=\large\frac{1}{x}$
Step 2:
Clearly when $f'(x) > 0$
$\Rightarrow \large \frac{1}{x}$$>0$ when $x >0$
Therefore $f'(x) > 0$
Hence $f(x)$ is an increasing function for $x > 0$
(i.e) $f(x)$ is increasing function whenever it is defined.