CAT Exam Model Paper 5 with solutions for free online practice

Question : 24 of 60

Marks: +1, -0

\log _ { 5 } \log _ { 4 } \log _ { 3 } ( x ^ { 2 } - 11 x + 1 ) = 0

(where x > 0) then what is the value of x?

Solution:

\log _ { 5 } \log _ { 4 } \log _ { 3 } ( x ^ { 2 } - 11 x + 1 ) = 0

\log _ { 4 } \log _ { 3 } ( x ^ { 2 } - 11 x + 1 ) = 1

\log _ { 3 } ( x ^ { 2 } - 11 x + 1 ) = 4

x ^ { 2 } - 11 x + 1 = 3 ^ { 4 } = 81

x ^ { 2 } - 11 x - 80 = 0

⇒

x ^ { 2 } - 16 x + 5 x - 80 = 0

⇒

x = 16

- 5

x = 16 (

given

x > 0 )

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