CAT Exam Model Paper 1 with solutions for free online practice

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Question : 86 of 100

Marks: +1, -0

If

\text"log"_{25}

5 = a and

\text"log"_{25}

15 = b, then the value of

\text"log"_{25}

27 is

3(b+a)
3(1 - b - a)
3(a + b-1)
3(1 - b + a)

Solution:

(c) Given,

\log _ { 25 } 15 = b

⇒\log _ { 25 } 5 + \log _ { 25 } 3 = b

⇒a + \log _ { 25 } 3 = b

⇒ \log _ { 25 } 3 = b - a

Now

\log _ { 25 } 5 = a

⇒ { \log 5 }/ { 2 \log 5 } = a

⇒a = { 1 }/ { 2 }

Again now,

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

= 3 ( b - a ) = 3 ( b - { 1 } /{ 2 } )

Going through the options, we have

3 ( a + b - 1 )

= 3 ( { 1 }/ { 2 } + b - 1 ) = 3 ( b - { 1 }/ { 2 } )

Hence, option (c) is correct.

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