# Indices (simple but i am have problems)

• January 8th 2012, 09:14 AM
kingmaths
Indices (simple but i am getting problems)
9^x+1 - 13(3^x) + 4 =0

Use y = 3^x

OK so i start and reached to this point
3^2(x+1) - 13(3^x) + 4 = 0

3^2x+2 - 13(3^x) + 4 = 0

I need some help to proceed, thanks in advance
• January 8th 2012, 09:34 AM
Bacterius
Re: Indices (simple but i am have problems)
$3^{2x + 2} - 13 \times 3^x + 4 = 0$

Expand:

$3^{2x} \times 3^2 - 13 \times 3^x + 4 = 0$

Which becomes (from the rules of indices - look at the first term):

$\left (3^{x} \right )^2 \times 3^2 - 13 \times 3^x + 4 = 0$

Now use the hint and substitute $y= 3^x$

$y^2 \times 3^2 - 13y + 4 = 0$

So:

$9 y^2 - 13y + 4 = 0$

Can you solve for $y$ and put the solution back in terms of $x$ now?