If \log_{(3x - 1)} (x - 2) = \log _{(9x^2 -6x +1)} (2x^2-10x - 2), then x equals -

KVPY SX SB Exam 2015 Question Paper

Question : 18 of 120

Marks: +1, -0

\log_{(3x - 1)} (x - 2) = \log _{(9x^2 -6x +1)} (2x^2-10x - 2),

then x equals -

Solution:

\log_{(3x - 1)} (x - 2) = \log _{( 3x-1)^2} (2x^2-10x - 2),

\log_{(3x - 1)} (x - 2)^2 = \log _{( 3x-1)} (2x^2-10x - 2),

(x - 2)^2 =2x^2- 10x - 2

x^ 2 - 4x + 4 = 2x^2- 10x - 2

x^ 2 - 6x - 6 = 0

x = 3 ± √15

x = 3 -√15\;\;\;x = 3 + √15

3 -√ 15

(x – 2) is negative

Go to Question:

Prev Question

Next Question