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UPSC NDA Math Model Paper 1

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Question : 28 of 120

Marks: +1, -0

If

\log 2, \log (2^{n}-1)

\log (2^{n}+3)

are in AP, then

n

is equal to:

$\log_{2} 5$
$\log_{3} 5$
5
$2^{5}$

Solution:

Since

\log 2, \log (2^{n}-1)

and

\log (2^{n}+3)

are in Arithmetic Progression, we can say that:

\log (2^{n}-1)-\log 2=\log (2^{n}+3)

-\log (2^{n}-1)

\Rightarrow \log \left( \frac{2^{n}-1}{2} \right)=\log \left( \frac{2^{n}+3}{2^{n}-1} \right)

\Rightarrow \frac{2^{n}-1}{2}=\frac{2^{n}+3}{2^{n}-1}

Let

2^{n}=x

\Rightarrow (x-1)^{2}=2(x+3)

\Rightarrow x^{2}-2 x+1=2 x+6

\Rightarrow x^{2}-4 x-5=0

\Rightarrow x^{2}-5 x+x-5=0

\Rightarrow x(x-5)+(x-5)=0

\Rightarrow (x-5)(x+1)=0

\Rightarrow x-5=0

OR

x+1=0

\Rightarrow x=5

OR

x=-1 .

\Rightarrow 2^{n}=5 \text{ OR } 2^{n}=-1 .

Since,

2^{n}=-1

is not possible, the only solution is

2^{n}=5 \Rightarrow \log_{2} 5=n

.

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