Question

If log 2, log (2^x − 1) and log (2^x + 3) are in A.P., write the value of x.

Answer 1

The numbers log 2, log (2^x − 1) and log (2^x + 3) are in A.P.

$$\begin{gathered} \therefore \log \left(2^x-1\right)-\log 2=\log \left(2^x+3\right)-\log \left(2^x-1\right) \\ \Rightarrow \log \left(\frac{2^x-1}{2}\right)=\log \left(\frac{2^x+3}{2^x-1}\right) \\ \Rightarrow \frac{2^x-1}{2}=\frac{2^x+3}{2^x-1} \\ \Rightarrow {\left(2^x-1\right)}^2=2\left(2^x+3\right) \\ \Rightarrow 2^{2x}+1-2.2^x=2.2^x+6 \\ \Rightarrow 2^{2x}-4.2^x-5=0 \end{gathered}$$

$$\begin{gathered} \mathrm{Let}{2}^x=y. \\ \Rightarrow y^2-4y-5=0 \\ \Rightarrow \left(y-5\right)\left(y+1\right)=0 \\ \Rightarrow y=5\mathrm{or}y=-1 \\ \therefore 2^x=5\mathrm{or}{2}^x=-1\left(\mathrm{not}\mathrm{possible}\right) \\ \mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{the}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ \log 2^x=\log 5 \\ \Rightarrow x\log 2=\log 5 \\ \Rightarrow x=\frac{\log 5}{\log 2}={\log }_{2}5 \\ \Rightarrow x={\log }_{2}5 \end{gathered}$$

Disclaimer: The question in the book has some error, so, this solution is not matching with the solution given in the book. This solution here is created according to the question given in the book.