13. The real numbers $a$ and $b$, where $a>b$, are solutions to the equation $3^{2 x}-10 \times 3^{x+1}+81=$ 0 . What is the value of $20 a^{2}+18 b^{2}$ ?
SOlUTION
198

In the equation $3^{2 x}-10 \times 3^{x+1}+81=0$, replace $3^{x}$ with $y$. The equation becomes $y^{2}-10 \times 3 \times y+81=0$. This factorises as $(y-3)(y-27)=0$ with solutions $y=3,27$. This means $3^{x}=3$ or $3^{x}=27$. The $x$-values are 1,3 respectively, so $a=3$ and $b=1$. The value of $20 a^{2}+18 b^{2}=20 \times 9+18 \times 1=198$.