Question

Solve the equation $9^{x+2}-6\times 3^{x+1}+1=0$

Answer 1

$9^{x+2}-6\times 3^{x+1}+1=0$

$9^2\times 9^x-6\times 3^1\times 3^x+1=0$

$81\times (3\times 3{)}^x-6\times 3\times 3^x+1$

$81\times 3^x\times 3^x-18\times 3^x+1=0$

Put $3^x=k$

$81\times k\times k-18\times k+1=0$

$81k^2-18k+1=0$

$81k^2-(9\times 9)k+1=0$

$81k^2-9k-9k+1=0$

$9k(9k-1)-1(9k-1)=0$

$(9k-1)(9k-1)=0$

$k=1/9$

$3^x=1/9=(3^2{)}^{-1}=3^{-2}$

$\therefore x=-2$.