Indian Institute of Foreign Trade 2012 Solved Paper

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Question : 24 of 128

Marks: +1, -0

If

\log3,\log(3^x-2)

and

\log(3^x+4)

are in arithmetic progression, then x is equal to

$8/3$
$\log_{3}8$
$\log_{2}3$
$8$

Solution:

If

\log3=\log(3^x-2)

and

\log(3^{x}+4)

are in arithmetic progression
Then,

2×\log(3^x-2)=\log3+\log(3^x+4)

Thus,

\log(3^{x}-2)^{2}=\log3(3^{x}+4)

Thus,

(3^x-2)^2=3(3^x+4)

⇒3^{2x}-4×3^x+4=3×3^x+12

⇒3^{2x}-7×3^x-8=0

⇒(3^x+1)×(3^x-8)=0

But

3^x+1≠0

Thus,

3^x=8

Hence

x=\log_{3}8

Hence, option B is the correct answer.

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