Question

Solve for X:

(3^2x-1) - 19(3^x) + 20 = 0

Answer 1

\[\dfrac{1}{3}\cdot3^{2x} - 19 \cdot 3^x + 20 = 0\]If you let \(t = 3^x\), then\[\dfrac{1}{3}t^2 - 19 t + 20 = 0\]

Answer 2

why the first term be 1/3?

Answer 3

\[3^{2x-1} = \dfrac{3^{2x}}{3^1} = \frac{1}{3}\cdot3^{2x}\]

Answer 4

Ohh thank you soo much!

Answer 5

No problem!

Answer 6

@ParthKohli Would you have to use the quadratic formula to find t then set it equal to 3x?

Answer 7

Exactly.

Ignore negative solutions, if any.

Answer 8

@ParthKohli So far I have \[1/2 ( 57+\sqrt{3009}) =3x\]

Answer 9

Yeah, that is one solution. But you shouldn't ignore the other one.

Answer 10

right then just divide both by 3?