Solve for x the equation 289^x - 17^(x+1) + 16 = 0?

We have to solve the equation 289^x - 17^(x+1) + 16 = 0.
Notice that 289 = 17^2.

Rewrite the equation that is given
in the following way to arrive at a quadratic
equation.

289^x - 17^(x+1) + 16 =
0

(17^2)^x - 17^(x+1) + 16 =
0

(17^x)^2 - 17^x*17 + 16 =
0

Let 17^x = y

y^2 - 17y + 16
= 0

y^2 - 16y - y + 16 = 0

y(y
- 16) - 1(y - 16) = 0

(y - 1)(y - 16) =
0

y = 1 and y = 16

As y =
17^x

17^x = 1 and 17^x = 16

x
= 0 and x =

The equation has two
solutions x = 0 and x =