# a 5÷2   b log5base2   c log5base3   d 3÷2

M manish

As we learned that in AP

$a_{2}-a_{1}=a_{3}-a_{2}$

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

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

$(2^{n}-1/2) = (2^{n}+3/2^{n}-1)$

Let $2^{n} = x$

$(x-1)^2 = 2x+6\Rightarrow x^2 -2x +1 = 2x +6$

$x^2 -4x-5=0\Rightarrow here$ X = 5 & X=-1 (X = -1 not possible)

So X = 5
$2^{n}= 5$ (Taking log both sides we get)

$n\log2 = \log5 \Rightarrow n = {log_{2}}^{5}$

Hence option (B) is correct

M manish

As we learned that in AP

$a_{2}-a_{1}=a_{3}-a_{2}$

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

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

$(2^{n}-1/2) = (2^{n}+3/2^{n}-1)$

Let $2^{n} = x$

$(x-1)^2 = 2x+6\Rightarrow x^2 -2x +1 = 2x +6$

$x^2 -4x-5=0\Rightarrow here$ X = 5 & X=-1 (X = -1 not possible)

So X = 5
$2^{n}= 5$ (Taking log both sides we get)

$n\log2 = \log5 \Rightarrow n = {log_{2}}^{5}$

Hence option (B) is correct

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