Math Problem

[supple bridge]

Okay, so I was bored and tried this problem out from the internet. I'm trying to get better at math, is this solution correct?

Problem: 3^x + 9^x = 27^x
Solution:
3^x + 9^x = 3^x + 3^x * 3^x = 3^x(3^x + 1)

3^x(3^x + 1) = 27^x
Divide both sides by 3^x
3^x + 1 = 9^x
3^x + 1 = (3^x)^2
(3^x)^2 - 3^x - 1 = 0
Let y = 3^x
y^2 - y - 1 = 0
y = (1 +- sqrt 5) / 2 = Φ
y = Φ
3^x = Φ
x = log_3 Φ

[tawny vine]

yes

but there is a simpler way tho

directly divide both sides by 27^x

(1/9)^x + (1/3)^x = 1

let y = (1/3)^x

y² + y = 1

y² + y - 1 = 0

y = (-1 ± sqrt(5))/2

(1/3)^x = (-1 ± sqrt(5))/2
x = log_1/3 (-1 ± sqrt(5))/2

[supple bridge]

that's pretty cool, but is my solution correct? is there a mistake in my solution?

[tawny vine]

no it's correct

[supple bridge]

ok cool thanks