Can $f(x+1) = f(x)^{\ln(x)}$ be expressed as integral transform $\int g(x,t) dt $?

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Let $x$ be a real number. Can some real-analytic function $f$ that satisfies for $x>3$ :$f(x+1) = f(x)^{\ln(x)}$ be expressed by standard functions as an integral transform :

$$f(x) = \int_0^{\infty} g(x,t) \space dt $$

Or maybe $$f(x) = \int_0^{\infty} \int_0^{\infty} h(x,t,r) \space dt \space dr$$

How fast does $f(x)$ grow ?

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Let $f(x)=a^{g(x)}$ , where $a\in\mathbb{R}^+$ and $a\neq1$ ,

Then $a^{g(x+1)}=(a^{g(x)})^{\ln(x)}$

$a^{g(x+1)}=a^{\ln(x)g(x)}$

$g(x+1)=\ln(x)g(x)$

$g(x)=\Theta(x)\prod\limits_x\ln(x)$, where $\Theta(x)$ is an arbitrary periodic function with unit period

$\therefore f(x)=a^{\Theta(x)\prod\limits_x\ln(x)}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period, $a\in\mathbb{R}^+$ and $a\neq1$