Solving for x in a Quadratic Exponential Equation

[theneedtoknow]

Homework Statement

3^2x - 12 * (3^x) + 3^3 = 0
solve for x

The Attempt at a Solution

I did this question just by factoring it into a more convenient form

3^x ( 3^x - 12) = -27
which tells me x>2 (in oder to make the bracket term, and hence the product, negative), and that x is a very small number (since the left side blows up very fast), so by logic i tried x=3 and it works

Of course, i don't like solving questions by guessing. Unfortunately I am having a brain fart today and can't come up with a way to explicitly solve for x using algebra (i've tried factoring, exponent rules, lns, but I can't figure it out)
How do I solve explicitly for x?

 

[rock.freak667]

Write 3^2x as (3^x)²

now put u=3^x and you now have a quadratic in 'u'.

 

[theneedtoknow]

haha oh man, i can't believe this stumped me so bad! Thank you so much! :)

 

[Mark44]

Homework Statement

3^2x - 12 * (3^x) + 3^3 = 0
solve for x

The Attempt at a Solution

I did this question just by factoring it into a more convenient form

3^x ( 3^x - 12) = -27

Sorry, but this is not a good idea. Just because two numbers multiply to a particular number doesn't let you say much of anything useful about the numbers. Unless the two numbers multiply to zero, though, then you know for certain that one or the other of the numbers must be zero.

This equation is quadratic in form, so go at the equation from that direction, as in factoring a trinomial.

which tells me x>2 (in oder to make the bracket term, and hence the product, negative), and that x is a very small number (since the left side blows up very fast), so by logic i tried x=3 and it works

Of course, i don't like solving questions by guessing. Unfortunately I am having a brain fart today and can't come up with a way to explicitly solve for x using algebra (i've tried factoring, exponent rules, lns, but I can't figure it out)
How do I solve explicitly for x?