Solve $3^{2x}-3^x\geq2$

Question

How to solve:

$$3^{2x}-3^x\geq2$$

I tried with $y=3^x$ and solved as equation: $y^2-y-2 \geq 0$

and I get:

$y<2$
$y>-1$

How should I proceed?

Answer 1

solutions you found are not correct. Correct solutions are $y\le -1$ and $y\ge 2$ but first solution gives no value of $x$ becuase $y$ is always positive. So now $3^x\ge 2$ gives $x\ge log_32$ which is the final answer.

Hope this helps !

Answer 2

When you have $ -1 > y$ or $ y > 2$,

replace $y$ with $3^x$, it gives you (remembering that $e > 0$): $$e^{x\ln(3)}>2$$ Which yields to: $$x > \dfrac{\ln(2)}{\ln(3)}$$

Answer 3

Note that if $u=3^x$, then $u^2=3^{2x}$. So in terms of $u$ we have $u^2-u\ge2$ or $u^2-u-2\ge0$. You can factor the quadratic polynomial $u^2-u-2$ to get a solution for $u$, then take logarithms to get it in terms of $x$.