How do i show that : for $x, y >0 ,\log (x+y)=\log{x}\cdot \log{y}$ has no integer solutions? I have tried to solve this equation for $x, y >0$, $\log(x+y)=\log{x}\cdot \log{y}$ by writing it as this $x+y={x}^{\log{y}}$. But the last expression is complicated to me to get any solutions, I think it has no solution in integers !!! 
My question here is : How do I show that for $x, y >0$,  $\log(x+y)=\log{x} \cdot \log{y}$ has no solution in integers?
Note: $x, y$ are integer numbers.
Edit 01: I edited the question because I have a wrong meaning, I meant in integers.
Edit 02: I edited the question to show my key idea for the proof.
Attempt: This is just an Idea for the proof 
What I think is  : $\log(x+y)\leq\log(xy)=\log{x}+\log{y}$ which is less than $\log{x} \cdot \log{y}$ for $ x, y \geq 8$, hence:
$2\leq \min(x,y)\leq 7$ , which  it's easy to check.
Thank you for any help !!!!
 A: The following expansion is close to your idea and assuming $\log{y}$ is the natural logarithm.
Case 1. Let's assume $x=y$ then $\ln{2y}=\ln^2{y}$. By replacing $\ln{y}=z$ it's easy to check that $P(z)=z^2-z-\ln{2}=0$ has 2 solutions, the largest one $z_2$ is in between $(1,\frac{7}{4})$ and $\forall z > z_2: P(z)>0$. It's easy to check that $z_2 < \frac{7}{4}$ from $\ln{2} < \frac{21}{16}$.
Case 2. Let's assume $x>y$ or $x=qy+r, 0\leq r<y, q>0$ then 
$$\ln((q+2)y) > \ln((q+1)y+r)=\ln(qy+r)\ln{y} \geq \ln(qy)\ln{y}$$
or
$$\ln(q+2) + \ln{y} > \ln{q}\ln{y} + \ln^2{y}$$
By replacing $\ln{y}=z$ 
$$P(z)=z^2+(\ln{q}-1)z-\ln(q+2)<0$$
Case 2.a $q=1$. This means
$$P(z)=z^2-z-\ln{3}<0$$
another easy to check case that the largest solution $z_2$ of $P(z)$ is in between $(1,\frac{7}{4})$ (from $\ln{3} < \frac{21}{16}$) and $\forall z > z_2: P(z)>0$.
Case 2.b $q\geq 2$. We will check the roots of $P(z)$
$$D=\sqrt{(\ln{q}-1)^2+4\ln(q+2)}$$
and
$$\sqrt{(\ln{q}-1)^2+4\ln{q}}<D<\sqrt{(\ln(q+2)-1)^2+4\ln(q+2)}$$
(including for $q=2$) or
$$\ln{q}+1<D<\ln(q+2)+1$$
And the largest of 2 solution for $P(z)$ is
$$z_2=\frac{-(\ln{n}-1)+D}{2}$$
or
$$1<z_2<\ln{\sqrt{1+\frac{2}{q}}}+1<\frac{7}{4}$$
and in fact, with large $q$, is aggressively approaching $1$. In any case, the largest solution $z_2$ is in between $(1,\frac{7}{4})$ and $\forall z > z_2: P(z)>0$.
Note: regarding $$\ln{\sqrt{1+\frac{2}{q}}}+1<\frac{7}{4}$$
For $q\geq 2$ we have $\frac{3}{4} > \frac{1}{2} \ln{2} \geq \ln{\sqrt{1+\frac{2}{q}}} \Rightarrow \frac{7}{4} > \ln{\sqrt{1+\frac{2}{q}}} +1$
In all the cases, we have $\ln{y}=z_2 \in (1,\frac{7}{4})$ or $y \in \{3, 4, 5\}$. The remaining part is to check each $\ln(x+3)=\ln{x}\cdot \ln{3}$, $\ln(x+4)=\ln{x}\cdot \ln{4}$ and $\ln(x+5)=\ln{x}\cdot \ln{5}$ individually, but numerical approaches reveal no integer solutions.
