Question

If $\log 2,\log (2^x-1),\log (2^x+3)$ are three consecutive terms of an $A.P$., then the value of $x$ will be

• A. ${\log }_{2}5$
• B. $0$
• C. ${\log }_{5}2$
• D. None of these

Answer 1

The correct option is A ${\log }_{2}5$

Since, $\log 2,\log (2^x-1),\log (2^x+3)$ are in A.P.
Therefore, $2\log (2^x-1)=\log 2+\log (2^x+3)\Rightarrow \log {(2^x-1)}^2=\log 2(2^x+3)\Rightarrow {(2^x-1)}^2=2(2^x+3)\Rightarrow {\left(2^x\right)}^2-4.2^x-5=0\Rightarrow (2^x-5)(2^x+1)=0\because 2^x\ne -1\therefore 2^x=5\Rightarrow x={\log }_{2}5$
Ans: A