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If ${\log }_{2}3=a/b$ were a rational number, choose positive integers $a,b$ because ${\log }_{2}3>0$. Exponentiation gives $2^a=3^b$, but the left side is even and the right side is odd. ThusThis is the basic irrational logarithm criterion: a rational logarithm would force positive integer powers of its base and argument to coincide.