<Dd> (2) log 2 ⁡ 3 = 3 . (\ displaystyle \ left ((\ sqrt (2)) \ right) ^ (\ log _ (\ sqrt (2)) 3) = 3 .) </Dd> <P> The base of the left side is irrational and the right side is rational, so one must prove that the exponent on the left side, log 2 ⁡ 3 (\ displaystyle \ log _ (\ sqrt (2)) 3), is irrational . This is so because, by the formula relating logarithms with different bases, </P> <Dl> <Dd> log 2 ⁡ 3 = log 2 ⁡ 3 log 2 ⁡ 2 = log 2 ⁡ 3 1 / 2 = 2 log 2 ⁡ 3 (\ displaystyle \ log _ (\ sqrt (2)) 3 = (\ frac (\ log _ (2) 3) (\ log _ (2) (\ sqrt (2)))) = (\ frac (\ log _ (2) 3) (1 / 2)) = 2 \ log _ (2) 3) </Dd> </Dl> <Dd> log 2 ⁡ 3 = log 2 ⁡ 3 log 2 ⁡ 2 = log 2 ⁡ 3 1 / 2 = 2 log 2 ⁡ 3 (\ displaystyle \ log _ (\ sqrt (2)) 3 = (\ frac (\ log _ (2) 3) (\ log _ (2) (\ sqrt (2)))) = (\ frac (\ log _ (2) 3) (1 / 2)) = 2 \ log _ (2) 3) </Dd>

Why every irrational number is a real number