Forgetting that the argument must be positivethe logarithm is not defined for zero or negative numbers.
Mixing up the two formslog2 8 = 3 because 23 = 8, not the other way round.
Solving incorrectly in exponential formif log3(x − 1) = 2 then x − 1 = 9, not 2 × 3.
Treating a logarithm as a productloga x is not equal to (loga) × x.