I'm assuming that you mean this:

$\displaystyle 3^{2x} + 3^{x + 1} = 0$

Now, $\displaystyle 3^{2x}$ is the same as $\displaystyle (3^x)^2$, and $\displaystyle 3^{x + 1}$ is the same as $\displaystyle 3(3^x)$ :

$\displaystyle (3^x)^2 + 3(3^x) = 0$

Now, let's substitute 3^x = t as javax suggested:

$\displaystyle t^2 + 3t = 0$

Hey, that looks like a quadratic equation!

Solve for t:

$\displaystyle t(t + 3) = 0$

$\displaystyle t = 0$ or $\displaystyle t = -3$

But we're not done! We need to solve for

**x**. So substitute 3^x back in for t:

$\displaystyle 3^x = 0$ or $\displaystyle 3^x = -3$

Hmm, there's

**no** x that would satisfy either equation. The base is positive and greater than 1, so the range of the function y = 3^x would be [0, ∞). So there is no solution, unless in the original equation the RHS equals some other number instead of 0.

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