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

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- Jan 8th 2012, 09:14 AMkingmathsIndices (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 - Jan 8th 2012, 09:34 AMBacteriusRe: Indices (simple but i am have problems)
$\displaystyle 3^{2x + 2} - 13 \times 3^x + 4 = 0$

Expand:

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

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

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

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

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

So:

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

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