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j4gym · Sep 20, 2009, 03:01 PM

How do you solve this problem?
3^2x+ 3^x -12 = 0
and can you do the steps for me to also please.

Perito · Sep 20, 2009, 08:11 PM

How do you solve this problem?
3^2x+ 3^x -12 = 0

It's fairly simple as soon as you realize you can factor it.

$3^{(2x)} + 3^x - 12 = 0$

$(3^x+4)(3^x-3)=0$

and can you do the steps for me to also please.

I wish I could (actually, I don't -- but I'm lazy). Just create two equations from the above equation. Move the constant to the right in each equation and take the logarithm of both sides.

Unknown008 · Sep 20, 2009, 11:22 PM

Okay, the easiest way to do this is by a simple substitution.

Let $y = 3^x$

Then,
$3^{2x} + 3^x-12 = 0$

$(3^x)^2 + 3^x-12 = 0$

$y^2 + y-12 = 0$

Now, you can solve it like any other quadratic;

$(y+4)(y-3) = 0$

Then,

y = -4or
y = 3

Replace y by 3^x

$3^x = -4$
$3^x = 3$

Since 3^x cannot be negative, it must be the other one.

$3^x = 3^1$

So, x = 1

:)