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