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Yes.

Take any rational number p.

Let a = any number that is not a power of 10, so that log(a) is irrational.

and let b = p/log(a).

log(a) is irrational so 1/log(a) must be irrational. That is, both log(a) and log(b) are irrational.

But log(a)*log(b) = log(a)*[p/log(a)] = p which is rational.

In the above case all logs are to base 10, but any other base can be used.

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