Question about inequality and logarithms

Question

Let's say we have an inequality:

$x < y$

does this hold true?

$\log x < \log y$

and if so, how can I prove it?

Answer 1

Let $\log_{B}{x}=a$ and $\log_{B}{y}=b$ for $B>0$. Then $ x=B^{a} $ and $ y=B^{b} $. So we can write $$x<y\Rightarrow B^{a}<B^{b}.$$ If $0<B\leq 1$, then

$$\begin{eqnarray} B^{a}<B^{b} & \Rightarrow & a>b \\ & \Rightarrow & \log_{B}{x}>\log_{B}{y} \end{eqnarray} $$but otherwise (say $B>1$),

$$\begin{eqnarray} B^{a}<B^{b} & \Rightarrow & a<b \\ & \Rightarrow & \log_{B}{x}<\log_{B}{y} \end{eqnarray}$$

Answer 2

Hint: $f(x)=\log_a (x)\Rightarrow f'(x)=\frac{1}{\ln (a) x}$