Derive $a^x$ using infinitesimal notation

Question

Consider $f(x) = a^x$. It holds that $$f' (x)\mathrm dx = f(x + \mathrm dx) - f(x)$$ using infinitesimal notation. Plugging all in, we get that $$f'(x)\mathrm dx = a^{x + \mathrm dx} - a^{x}.$$ Is it true that $a^{x+\mathrm dx} - a^x = a^{x}(a^{\mathrm dx} - 1)$ using algebra? And if yes, how can I justify that $a^{\mathrm dx} - 1 = \log(a)$? This should hold since $f'(x) = a^x\log(a)$ using the rgular $\epsilon$-$\delta$ derivative.