Equations with variable powers Find the roots of the equation $3^{x+2}$+$3^{-x}$=10 . By inspection the roots are $x=0$ and $x=-2$. But how can I solve this equation otherwise? 
 A: Suppose $y=3^x$. Then $3^{x+2}$+$3^{-x}=10$ is transformed into $$9y+{1\over y}=10\Rightarrow 9y^2-10y+1=0\Leftrightarrow(y-1)(9y-1)=0$$
A: Hint: Multiply both sides by the positive quantity $3^x$ to clear the fraction implied by the negative exponent, and rearrange a bit:
$$3^{x+2} + 3^{-x}=10$$
$$3^x\cdot3^{x+2} + 3^x\cdot 3^{-x} = 3^x\cdot 10$$
$$3^{2x+2} + 1 = 3^x\cdot 10$$
$$3^{2x}\cdot 3^2 + 1 = 3^x\cdot 10$$
$$(3^x)^2\cdot 9 + 1 = 3^x\cdot 10$$
$$9\cdot(3^x)^2 - 10\cdot(3^x) + 1 = 0$$
Now think of this as a quadratic equation in $3^x$:
$$9u^2 - 10u + 1 = 0$$
where $u = 3^x$.
Solve the quadratic to get the value(s) of $u$, say $u=A$ and $u=B$. Then remember that $u$ stands for $3^x$, so now you have to solve the equations $3^x=A$ and $3^x=B$, where $A$ and $B$ are the solutions to your quadratic.
Note that you will only have real solutions if $A>0$ or $B>0$, since $3^x$ can never be negative or zero if $x$ is real.
A: your equation is equivalent to $3^{2x}-\frac{10}{9}3^x+\frac{1}{9}=0$ solve this as a quadratic equation.
