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

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.

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

:)