Irrationality of sum of two logarithms: $\log_2 5 +\log_3 5$

I try to prove that the number $$\log_2 5 +\log_3 5$$ is irrational. But I have no idea how to do it. Any hints are welcome.

Solution 1:

Questions of this form fall under the general heading of transcendental number theory. Conditional on a major conjecture in this field called Schanuel's conjecture, this number is not only irrational but transcendental. In fact, conditional on Schanuel's conjecture much more is true:

(Conditional) Theorem: The logarithms $\log 2, \log 3, \log 5, \dots$ of the primes are algebraically independent.

Essentially this means that any rational function of the logarithms of the primes is transcendental unless you can simplify it, as a rational function, to a rational number. The conclusion follows because your expression can be written $\frac{\log 5}{\log 2} + \frac{\log 5}{\log 3}$.

Proof. By unique prime factorization, the logarithms of the primes are linearly independent over $\mathbb{Q}$. If $p_1, p_2, \dots$ is an enumeration of the primes, then by Schanuel's conjecture it follows that $\mathbb{Q}(\log p_1, \log p_2, \dots \log p_k)$ has transcendence degree at least $k$, hence exactly $k$, for all $k$. $\Box$

Solution 2:

If both of them are rational then...

$$ \log_2(5)=\frac{m_1}{n_1} \implies 2^{m_1/n_1}= 5 \implies 2^{m_1}=5^{n_1} $$ $$ \log_3(5)=\frac{m_2}{n_2}\implies 3^{m_2/n_2}=5 \implies 3^{m_2}=5^{n_2} $$

The rightmost equation would only be true if both $n,m$ are $0$, but that contradicts the leftmost equation ($n_{1/2}\in\mathbb{N_{/0}},m_{1/2}\in\mathbb{Z}$)

For $n,m > 1$ the rightmost line can't be true because it contradicts prime factorization. (due to this fact we can also ignore the following cases: $0<n_{1/2}<1<m_{1/2};0<n_{1/2},m_{1/2}<1; 0<m_{1/2}<1<n_{1/2}$

So the sum of two irrational numbers ($i_1, i_2$) is rational if $i_1+i_2=\frac{m}{n} $ for $\frac{m}{n}$ being the rational sum.

$$ \frac{m}{n}=\log_3(5)+\log_2(5) \implies 2^{m/n}=2^{\log_3(5)+\log_2(5)}\implies $$ $$ 2^{m/n}=5\times2^{\log_3(5)}\implies 2^m=5^{n}\times2^{n\log_3(5)} $$

And here is where we have the next contradiction: The left side is always rational but $2^{n\log_3(5)}$ is irrational and so is the product on the right side of the last equation (because the product of a rational and irrational number is always irrational).

Therefore the statement is false and the sum has to be irrational. That was a tricky one!