Question

If ${\log }_{10}2,{\log }_{10}(2^x-1)and{\log }_{10}(2^x+3)$ be three consecutive terms of an A.P., then

• A. $x=0$
• B. $x=1$
• C. $x={\log }_{2}5$
• D. $x={\log }_{10}2$

Answer 1

Answer: (C)

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

Since the three terms are in A.P.
$\therefore 2{\log }_{10}(2^x-1)={\log }_{10}2+{\log }_{10}(2^x+3)$ $[2b=c+a]$
or ${(2^x-1)}^2=2(2^x+3)$ (using logarithmic properties)

or ${(y-1)}^2=2(y+3)wherey=2^x$
or $y^2-4y-5=0$
$\therefore (y-5)(y+1)=0ory=5,-1$
$\because 2^x=y\ne -1,$ Exponential function can't be -ve
$\Rightarrow 2^x=5\therefore x={\log }_{2}5.$