Question

If $\log \left(2\right),\log \left(2^n-1\right)$ and $\log \left(2^n+3\right)$ are in A.P. then $n=$

• A. $\frac{5}{2}$
• B. ${\log }_2\left(5\right)$
• C. ${\log }_3\left(5\right)$
• D. $\frac{3}{2}$

Answer 1

The correct option is B

${\log }_2\left(5\right)$

Explanation for the correct answer:

Step 1: Simplify the given equation using log properties:

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

If $a,b,c$ are in A.P then $2b=a+c$

$$\begin{gathered} \Rightarrow 2\log \left(2^n-1\right)=\log \left(2\right)+\log (2^n+3) \\ \Rightarrow 2\log \left(2^n-1\right)=\log \left(2\right(2^n+3\left)\right)\left[\because logab=loga+logb\right] \\ \Rightarrow \log {\left(2^n-1\right)}^2=\log \left(2\right(2^n+3\left)\right)\left[\because blog\left(a\right)=log{\left(a\right)}^b\right] \\ \Rightarrow {\left(2^n-1\right)}^2=\left(2\right(2^n+3) \\ \Rightarrow 2^{2n}+1-2·2^n=2·2^n+6\left[\because {\left(a-b\right)}^2=a^2+b^2-2ab\right] \\ \Rightarrow 2^{2n}+1-2^{n+1}=2^{n+1}+6 \\ \Rightarrow 2^{2n}-2^{n+2}-5=0 \end{gathered}$$

Step 2: Factorize the equation to get the value of $x$:

Let $2^n=y$

$$\begin{gathered} \Rightarrow y^2-4y-5=0 \\ \Rightarrow y^2-5y+y-5=0 \\ \Rightarrow y(y-5)+1(y-5)=0 \\ \Rightarrow (y+1)(y-5)=0 \\ \Rightarrow y=-1,5 \end{gathered}$$

So $2^n=5\Rightarrow n={\log }_2\left(5\right)$ $\left[-\mathrm{ve}\mathrm{value}\mathrm{not}\mathrm{possible}\right]$

Hence option (B) i.e. ${\log }_2\left(5\right)$ is correct.