Question

if $\text{log}\text{2,}\text{log}\left({\text{2}}^{\text{n}}\text{- 1}\right)\text{and}\text{log}\left({\text{2}}^{\text{n}}\text{+ 3}\right)$ are in $\text{AP,}$ then $n$ is equal to

• A. $\frac{5}{2}$
• B. ${\log }_{2}5$
• C. ${\log }_{3}5$
• D. $\frac{3}{2}$

Answer 1

Answer: (B)

${\log }_{2}5$

Given $\text{log}\text{2,}\text{log}\left({\text{2}}^{\text{n}}\text{- 1}\right)\text{and}\text{log}\left({\text{2}}^{\text{n}}\text{+ 3}\right)$ are in $AP$.

Hence,

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

$⟹(2^n-1{)}^2=2^{n+1}+6$

$⟹2^{2n}+1-2^{n+1}=2^{n+1}+6$

$⟹2^{2n}-2^{n+2}-5=0$

Now, let $2^n=y$. Then,

$⟹y^2-4y-5=0$

So,

$y=5$..........$[y=-1(reject\left)\right]$

Hence,

$2^n=5⟹n={\log }_{2}5$