How to Solve the Equation 2^x^2-3=4^x?

[star321]

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

i can't 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, I am sure i can figure it out from there.
thankyou

 

[AKG]

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

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

It's a polynomial in x.

2^x2 - 3 = 4^x...

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

 

[0rthodontist]

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

 

[star321]

still a bit confused

it was the first one, the one you were working out.
since your multiplying ln2 by (x^2-3) wouldn't 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 doesn't work out, what am I doing wrong?

 

[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

 

[GregA]

just to expand on what's been said and offer another way of starting off...
2^x2-3 = 4^x is a bit like saying
2^x2-3 = 2^2x
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