Question

If $\log 2,\log (2^x-1),\log (2^x+3)$ are in A.P., then $x$ is

• A. $1$
• B. $\frac{3}{2}$
• C. ${\log }_{2}5$
• D. ${\log }_{2}3$

Answer 1

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

If $\log 2,\log (2^x-1),\log (2^x-3)$ are in $A.P$ then $x$

Since the term are in $A.P$

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

let $2^x=6$

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

$2\log (t-1)=\log 2(t+3)$

$(t-1{)}^1=2(t+3)$

$t^2+1-2t=2t+6$

$t^2+1-2t=2t+6$

$t^2-4t-5=0$

$(t-5)(t+1)=0$

$t=5,-1$

out $a^x\ne -1\therefore 2^x=5$

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