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Howard C. Saar of Albion, Mich., pointed out an innovative solution to this problem in Recreational Mathematics Magazine, April 1962:

log(3x + 2) + log(4x – 1) = 2log11

Divide each side of the equation by the word “log”:

(3x + 2) + (4x – 1) = (2)(11)

7x = 21

x = 3

… which is correct.