Show that $\log_a x$ (for some $x > 1$), as a function of $a$, is monotonically decreasing

Question

Let $a>0$. Let define logarithm function with base $a$ as a inverse function of $f(x)=a^x$ i.e. $log_a x = f^{-1}(x)$. Next we fix $x>1$ and define $g(a)=log_a x$. Show that function $g(a)$ is monotonically decreasing.

How I should answer this?

Answer 1

If $b>a>1$, then $a^{\log_ax}=x$, whereas$$a^{\log_bx}=\left(\frac ab\right)^{\log_bx}b^{\log_bx}=\left(\frac ab\right)^{\log_bx}x<x,$$since $\frac ab<1$. Therefore, $\log_bx<\log_ax$.

Answer 2

Let $y >0$ and consider

$f(y)= \log_y (c)$ , where

$c$ >1 is a constant.

$z:= \log_y (c)$ then $y^z = c$;

$z \ln y = \ln c$, or $z= \dfrac{\ln c}{\ln y}$.

With $f(y) = \dfrac{\ln c}{\ln y}$ we get:

$f'(y) =(-1) (\ln c)(\ln y)^{-2}(1/y).$

Since $c>1$ we have $\ln c >0$, and

$f'(y)<0$, hence decreasing