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 proove $x=lo{g}_{2}5$.

Answer 1

Given: ${\log }_{10}2,$ ${\log }_{10}(2^x-1)$ and ${\log }_{10}(2^x+3)$ are in A.P.

Therefore,

$⟹2{\log }_{10}(2^x-1)={\log }_{10}2+{\log }_{10}(2^x+3)⟹2{\log }_{10}(2^x-1)={\log }_{10}2(2^x+3)⟹{\log }_{10}{(2^x-1)}^2={\log }_{10}2(2^x+3)⟹{(2^x-1)}^2=2(2^x+3)⟹{(y-1)}^2=2(y+3)[Lety=2^x]⟹y^2-2y+1=2y+6⟹y^2-2y-2y=6-1⟹y^2-4y-5=0⟹y^2-5y+y-5=0⟹y(y-5)+1(y-5)=0⟹(y-5)(y+1)=0\therefore y=5ory=-1$

But $y=2^x\ne -1$ [Exponential function is not negative.]

Thus, $y=2^x=5$

$\therefore x={\log }_{2}5$ [Hence Proved]