NIMCET 2014 Question Paper with solutions

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Question : 40 of 120

Marks: +1, -0

If a =

\log_{12}

18 , b =

\log_{24}

54 , then ab + 5 (a - b) is

1
0
2
$3/2$

Solution:

We have,
a =

\log_{12}

18
⇒ b =

\log_{24}

54
∴ ab + 5 (a - b)
= (

\log_{12}

18) (

\log_{24}

54) + 5 [

\log_{12}

18 -

\log_{24}

54]
=

{\log_e 9×2}/{\log_e 4×3}

.

{\log_e 9×6}/{\log_e 8×3}

+ 5

[{\log_e 9×2}/{\log_e 4×3}-{\log_e 9×6}/{\log_e 8×3}]

Solving this, we get
ab + 5(a-b) = 1

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