Solve 5^2x – 6 (5^x) + 5 = 0

Please help me solve this
Thanks!

Lets give 5^x a value of y:

y^2 - 6y + 5 = 0

Factor it out:

(y - 5) (y - 1) = 0

Solve for each:

y - 5 = 0
y = 5

y - 1 = 0
y = 1

Now substitute 5^x back for y and solve for each:

5^x = 5
x = 1

5^x = 1
x = 0
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Overall:

x = 0, 1

5^(2x) - [6 * 5^(x)] + 5 = 0 → you know that: x^(ab) = x^(ba)

5^(x * 2) - [6 * 5^(x)] + 5 = 0 → you know that: x^(a * b) = (x^a)^b

[5^(x)]² - 6 * 5^(x) + 5 = 0 → let: X = 5^x → where X > 0 of course

X² - 6X + 5 = 0

Polynomial like: ax² + bx + c, where:
a = 1
b = - 6
c = 5

Δ = b² - 4ac (discriminant)

Δ = (- 6)² - 4(1 * 5) = 36 - 20 = 16 = 4²

X1 = (- b - √Δ) / 2a = (6 - 4) / (2 * 1) = 1

X2 = (- b + √Δ) / 2a = (6 + 4) / (2 * 1) = 5


First case: 5^x = 1

Ln(5^x) = Ln(1)

x.Ln(5) = 0

→ x = 0


Second case: 5^x = 5

5^x = 5^1

→ x = 1

Solution = { 0 ; 1 }