hello i am trying to solve this equation for the values of x. can i get some help solving this one.
$
3^{2x-1} -4(3^x) + 1= 0
$

2. Originally Posted by sigma1
hello i am trying to solve this equation for the values of x. can i get some help solving this one.
$
3^{2x-1} -4(3^x) + 1= 0
$
Hi sigma1,

$3^{2x-1} -4(3^x) + 1= 0$

$3^{-1}3^{2x} -4(3^x) + 1= 0$

$\left(\frac{1}{3}\right)\left(3^{x}\right)^2 -4(3^x) + 1= 0$

$\left(\frac{1}{3}\right)y^2 -4y + 1= 0$

$y^2-12y+3=0$

where $y = 3^x$.

3. Originally Posted by undefined
Hi sigma1,

$3^{2x-1} -4(3^x) + 1= 0$

$3^{-1}3^{2x} -4(3^x) + 1= 0$

$\left(\frac{1}{3}\right)\left(3^{x}\right)^2 -4(3^x) + 1= 0$

$\left(\frac{1}{3}\right)y^2 -4y + 1= 0$

$y^2-12y+3=0$

where $y = 3^x$.
thanks alot.. i finally have some perspective on this one