Question

If $lo{g}_{3}2,lo{g}_3(2^x-5)and$ $lo{g}_3(2^x-\frac{7}{2})$ are in A.P, then x is equal to

Answer 1

$lo{g}_3^2,lo{g}_3^{(2^x-5)}$ and $lo{g}_3^{(2^x-\frac{7}{2})}$ are in A.P

$\Rightarrow 2lo{g}_3^{(2^x-5)}=lo{g}_3^2+lo{g}_3^{(2^x-\frac{7}{2})}$

$\Rightarrow lo{g}_3^{(2^x-5{)}^2}=lo{g}_3^{2(2^x-7/2)}$

$\Rightarrow (2^x-5{)}^2=2(2^x-7/2)$

$\Rightarrow 2^{2x}-102^x+25={2.2}^x-7$

$\Rightarrow 2^{2x}-{12.2}^x+32=0$

$\Rightarrow 2^{2x}-{8.2}^x-{4.2}^x+32=0$

$\Rightarrow 2^x(2^x-8)-4(2^x-8)=0$

$(2^x-4)(2^x-8)=0$

$\Rightarrow 2^x=4$ or $2^x=8$

$2^x=4$ is not possible $\therefore 2^x=8=2^3$

as $lo{g}_2^{(2^x-5)}$ will not be defined. $\Rightarrow x=3$.