Question

Q: If log2,log(2^n-1) and log (2^n +3) are in AP ,then n=

Answer 1

this may help you my friend.

Answer 2

Answer:

The value of n is $n=\log_2(5)$

Step-by-step explanation:

Given : If $\log2,\log(2^n-1),\log (2^n +3)$ are in A.P.

To find : The value of n?

Solution :

We know that, In an A.p the common difference between two term is same.

So, $a_2-a_1=a_3-a_2$

Where, $a_1=\log 2,a_2=\log(2^n-1),a_3=\log (2^n +3)$

$\log(2^n-1)-\log2=\log(2^n+3)-\log(2^n-1)$

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

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

$(2^n-1)(2^n-1)=(2^n+3)(2)$

$(2^n-1)^2=2^{n+1}+6$

$2^{2n}+1-2^{n+1}=2^{n+1}+6$

$2^{2n}=2^{n+2}+5$

$2^{2n}-2^{n+2}-5=0$

Let $2^n=x$

$x^2-4y-5=0$

Applying middle term split,

$(x-5)(x+1)=0$

$x=5,x=-1$

Reject x=-1.

$2^n=5$

$n=\log_2(5)$

Therefore, The value of n is $n=\log_2(5)$