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.