Question

If $lo{g}_{10}2,lo{g}_{10}(2^x-1)$ and $lo{g}_{10}(2^x+3)$ be three consecutive terms of an A.P., then $x=lo{g}_{2}5$.

Answer 1

Since the three terms are in A.P.
$\therefore 2lo{g}_{10}(2^x-1)=lo{g}_{10}2+lo{g}_{10}(2^x+3)$
or $(2^x-1{)}^2=2(2^x+3)$
or $(y-1{)}^2=2(y+3)$ where $y=2^x$
or $y^2-4y-5=0$
$\therefore (y-5)(y+1)=0$ or $y=5$,
$\because 2^x=y\ne -1$. Exponential $f^n$ is not $-$ive
or $2^x=5$
$\therefore x=lo{g}_{2}5$.