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, 10:14 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, 10:34 AMBacteriusRe: Indices (simple but i am have problems)

Expand:

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

Now use the hint and substitute

So:

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