What's the monotony of this function?

Question

This is the function: $g(x) = (1+a)^x - a^x$, for some $a>0$ and $x \ge 0$

I can find the monotony for $1>a>0$ this way. Let $x_1, x_2$ be two non-negative numbers such that: $$x_1<x_2 \iff (1+a)^{x_1} < (1+a)^{x_2} \tag{1}$$

$$x_1 < x_2 \iff a^{x_1} > a^{x_2} \iff -a^{x_1} < -a^{x_2} \tag{2}$$

So then: $(1) + (2) \implies g(x_1) < g(x_2)$.

But what about $a\ge 1$?

Answer 1

$$(1+a)^x-a^x=a^x\left(\left(1+\frac1a\right)^x-1\right).$$

For $a>1$ and $x>0$, both factors are positive growing functions.

Answer 2

As Stefan4024 commented, using derivatives makes life easy.

Considering the function $$f(x)=(1+a)^x-a^x$$ $$f'(x)=(a+1)^x \log (a+1)-a^x \log (a)$$ The first derivative cancels at $$x_*=\frac{\log \left(\frac{\log (a)}{\log (a+1)}\right)}{\log (a+1)-\log (a)}$$ which is a negative number for any $a >0$. So, for any $x>0$, $f'x)>0$ and $f(x)$ is an increasing function.