Indian Institute of Foreign Trade 2015 Solved Paper

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Question : 4 of 124

Marks: +1, -0

If log

_25

5 = a and log

_25

15 = b, then the value of log

_25

27 is :

3(b + a)
3 (1 – b –a)
3 (a + b –1)
3 (1 – b + a)

Solution:

(c)

\log _ { 25 } 27 = \log _ { 25 } 9 × 3

= 3 \log _ { 25 } 3

\;\;\;\;\;\;\;\;\;\;\;...(1)

Now,

a + b = \log _ { 25 } 5 + \log _ { 25 } 15

= \log _ { 25 } ( 5 × 15 ) = \log _ { 25 } 75

= \log _ { 25 } ( 25 × 3 )

= \log _ { 25 } 25 + \log _ { 25 } 3

= 1 + \log _ { 25 } 3

\log _ { 25 } 3 = a + b - 1\;\;\;\;\;\;\;...(2)

∴ \log _ { 25 } 27 = 3 ( a + b - 1 )

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