Question

How to solve 3^2x=7(3^x)-12

Answer 1

I'm looking for some oranges and a damp handkerchief, where would would I go for this?

Answer 2

\[3^{2x} = 7*3^x - 12 \iff \\(3^x)^2 - 7*3^x + 12 = 0\]

Answer 3

you now have a quadratic in 3^x, in that, if you think of 3^x as being y, then you have something that looks like:
\[3^x = y, then\\ y^2 - 7y + 12 = 0\]

Answer 4

You can solve that quadratic quite easily, to get two values for y (i.e. two values for 3^x). Now use those to find x.

Answer 5

how come when you take out the negative from the equation you don't put it like -(y^2-7y+12)=0

Answer 6

because if -(y^2 - 7y + 12) = 0, then y^2 - 7y + 12 = 0.

Think about this: does 0 = -0 ? It does (it's the only number that does that, actually). And you say that - (y^2 - 7y + 12) = 0. So then you can write - (7^2 - 7y + 12) = 0 = -0 = y^2 - 7y + 12.

Another way of looking at it, is that if you have two numbers a, b, and a = b, then for any other number c, c*a = c*b. Here a = - (y^2 - 7y + 12) and b = 0, and c = -1.

The other thing is that I didn't even think of those properties when I derived my expression. I just moved everything to the left:
y^2 = 7y - 12. Then add -(7y-12) to both sides to get: y^2 - 7y + 12 = 0.