# Solve the equation for x

1. Jun 26, 2006

### star321

solve the equation:
2^x^2-3=4^x (2 to the power of x squared - 3)

i cant get it to work
ln2^x^2 - 3=ln4^x
(x^2 -3) ln (2)=xln(4)
i know it'd be -3(ln2)=xln4... but i don't know what to do with x^2 part.

i did other questions like this but with 5^2x and 3^1+x, what do I do with the X^2 that gets me stuck. If someone knows what I do with taht could you please tell me, im sure i can figure it out from there.
thankyou

2. Jun 26, 2006

### AKG

Do you mean 2x2-3 = 4x or 2x2 - 3 = 4x?

2x2-3 = 4x
(x2 - 3)ln(2) = xln(4)
ln(2)x2 - ln(4)x - 2ln(3) = 0

It's a polynomial in x.

2x2 - 3 = 4x...

Well actually, do you mean 2(x2) - 3 ... or do you mean (2x)2 - 3 ...?

3. Jun 26, 2006

### 0rthodontist

Probably a typo, the constant term in the polynomial is just -3.

4. Jun 26, 2006

### star321

still a bit confused

it was the first one, the one you were working out.
since your multiplying ln2 by (x^2-3) wouldnt it be 3 ln2?
So from ln(2)x2 - ln(4)x - 2ln(3) = 0
you get (x-6)(x+2) so x=6, x=-2.
I tried putting this numbers in the original but it doesnt work out, what am I doing wrong?

5. Jun 26, 2006

### AKG

Oh yeah, 3ln(2). So you'd get:

ln(2)x2 - ln(4)x - 3ln(2) = 0
x2 - 2x - 3 = 0
(x-3)(x+1) = 0

6. Jun 26, 2006

### GregA

just to expand on what's been said and offer another way of starting off...
2x2-3 = 4x is a bit like saying
2x2-3 = 22x
2 raised to the power on the LHS is equal to 2 raised to the power on the RHS such that both exponents are the same, ie: x2-3 = 2x

Last edited: Jun 26, 2006