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Log2,log(2^n-1),log(2^n+3) in AP, then n=?

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

Answers (2)
9 Views
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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