Exams › JEE Main › Maths › Logarithms and Their Properties
1 questions with worked solutions.
Answer: 2
Setting log₉(a) = log₁₅(b) = log₂₅(2a+b) = k gives a=9^k, b=15^k, 2a+b=25^k. Substituting: 2*9^k + 15^k = 25^k. Let t = (5/3)^k; dividing by 9^k gives 2 + t = t², so t² - t - 2 = 0, giving t = 2 (positive). Then b/a = 15^k/9^k = (5/3)^k = t = 2.