Question

If $log2,log(2^x-1)andlog(2^x+3)$ are in A.P., then x is equal to :

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

Answer 1

Answer: (B)

${log}_{2}5$

$\log 2,\log (2^x-1),\log (2^x-3)$ are in AP

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

${\log (2^x-1)}^2=\log \left[2.(2^x+3)\right]$

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

${\left(2^x\right)}^2+1-2.2^x=2.2^x+6$

${\left(2^x\right)}^2-4.2^x-5=0\Rightarrow {\left(2^x\right)}^2-5.2^x+2^x-5=0\Rightarrow 2^x(2^x-5)+1(2^x-5)=0\Rightarrow (2^x-5)(2^x+1)=0$

$2^x+1\ne 0,2^x+5=0\Rightarrow 2^x=5$

taking log of base 2

$x{\log }_{2}2={\log }_{2}5$

$x={\log }_{2}5$