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As Stinking Bishop mentions in the comments, if $f(a)+f(b)=f(a^b)$, then it follows (by interchanging $a$ and $b$) that $f(b)+f(a)=f(b^a)$. Hence, $f(a^b)=f(b^a)$ for all $a,b\in\mathbb R^+$. Setting $b=1$, we see that $f(a)=f(1)$ for all $a\in\mathbb R^+$. Now $f(1)+f(1)=f(1^1)$, so $f(1)=0$. Hence, the only function $\mathbb R^+\to\mathbb R$ with the desired property is the zero function.

If $f(a)+f(b)=f(a^b)$ for all $a,b\in\mathbb R^+$, then it follows (by interchanging $a$ and $b$) that $f(b)+f(a)=f(b^a)$ for all $a,b\in\mathbb R^+$. Hence, $f(a^b)=f(b^a)$ for all $a,b\in\mathbb R^+$. Setting $b=1$, we see that $f(a)=f(1)$ for all $a\in\mathbb R^+$. Now $f(1)+f(1)=f(1^1)$, so $f(1)=0$. Hence, the only function $\mathbb R^+\to\mathbb R$ with the desired property is the zero function.

As Stinking Bishop mentions in the comments, if $f(a)+f(b)=f(a^b)$, then it follows (by swapping $a$ and $b$) that $f(b)+f(a)=f(b^a)$. Hence, $f(a^b)=f(b^a)$ for all $a,b\in\mathbb R^+$. Setting $b=1$, we see that $f(a)=f(1)$ for all $a\in\mathbb R^+$, so $f$ is a constant function. Now $f(1)+f(1)=f(1^1)$, so $f(1)=0$. Hence, the only function $\mathbb R^+\to\mathbb R$ that satisfies your requirements is the zero function.

As Stinking Bishop mentions in the comments, if $f(a)+f(b)=f(a^b)$, then it follows (by interchanging $a$ and $b$) that $f(b)+f(a)=f(b^a)$. Hence, $f(a^b)=f(b^a)$ for all $a,b\in\mathbb R^+$. Setting $b=1$, we see that $f(a)=f(1)$ for all $a\in\mathbb R^+$. Now $f(1)+f(1)=f(1^1)$, so $f(1)=0$. Hence, the only function $\mathbb R^+\to\mathbb R$ with the desired property is the zero function.

As Stinking Bishop mentions in the comments, if $f(a)+f(b)=f(a^b)$, then it follows (by swapping $a$ and $b$) that $f(b)+f(a)=f(b^a)$. Hence, $f(a^b)=f(b^a)$ for all $a,b\in\mathbb R^+$. Setting $b=1$, we see that $f(a)=f(1)$ for all $a\in\mathbb R^+$, so $f$ is a constant function. Now $f(1)+f(1)=f(1^1)$, so $f(1)=0$. Hence, the only function that satisfies your requirements is the zero function.

As Stinking Bishop mentions in the comments, if $f(a)+f(b)=f(a^b)$, then it follows (by swapping $a$ and $b$) that $f(b)+f(a)=f(b^a)$. Hence, $f(a^b)=f(b^a)$ for all $a,b\in\mathbb R^+$. Setting $b=1$, we see that $f(a)=f(1)$ for all $a\in\mathbb R^+$, so $f$ is a constant function. Now $f(1)+f(1)=f(1^1)$, so $f(1)=0$. Hence, the only function $\mathbb R^+\to\mathbb R$ that satisfies your requirements is the zero function.