Is this sum of 2 different irrational logarithms irrational: $\log_2(3)+\log_3(2)$? 
I am having some problems proving that the following sum is irrational or rational:
$$\log_2(3)+\log_3(2)$$

This is all I've got for now:
$\log_2(3)=\frac mn \iff 2^{\frac mn}=3 \iff 2^m=3^n$ so $\log_2(3)$ = irrational.
$\log_3(2)=\frac qr \iff 3^{\frac qr}=2 \iff 3^q=2^r$ so $\log_3(2)$ = irrational.
Now I'm having trouble with proving that $\log_2(3)+\log_3(2)$ is irrational. I know that the sum of two irrational numbers isn't directly irrational. Also, both base numbers of the logarithms are primes.

Let $x=\log_23$. Note that $\log_32=\frac1x$, so if $x+\frac1x$ were rational we would have $x+\frac1x=q\in\mathbb Q$, or: $x^2-qx+1=0$, meaning that $x$ is an algebraic number (of degree at most $2$). Now we can rewrite $\log_23=x$ a bit and use a strong theorem, the Gelfond-Scheider theorem, to obtain a contradiction:
$$2^x=3.$$
Gelfond-Schneider says
$$(\text{algebraic and not 0 or 1})^{\text{algebraic and irrational}}=\text{transcendental}.$$
Because $2$ is algebraic and not $0$ or $1$, $x$ is algebraic and irrational but $3$ is not transcendental, we obtain a contradiction.
Note: Using the same method we can prove that $\log_23+\log_32$ is in fact transcendental; this is because the field (yes, it's a field, and that's not trivial) of algebraic numbers (contained in $\mathbb C$) over $\mathbb Q$ is algebraically closed.
