STANFORD UNIVERSITY Admission Interview Secrets Revealed!

[@key_board_x]

log(x) + log(y) = 5 ← this is the sum S
log(x).log(y) = 5 ← this is the product P
log(x) & log(y) are the solution of the following equation:
z² - Sz + P = 0
z² - 5z + 5 = 0
Δ = (- 5)² - (4 * 5) = 25 - 20 = 5
z = (5 ± √5)/2
log(x) = (5 + √5)/2 and log(y) = (5 - √5)/2
or
log(x) = (5 - √5)/2 and log(y) = (5 + √5)/2
Recall: log(a) = Ln(a) / Ln(10)
Recall: Ln(a) = x → a = e^(x)
First case: log(x) = (5 + √5)/2 and log(y) = (5 - √5)/2
log(x) = (5 + √5)/2
Ln(x)/Ln(10) = (5 + √5)/2
Ln(x) = [(5 + √5)/2].Ln(10)
Ln(x) = Ln(10)^[(5 + √5)/2]
→ x = 10^[(5 + √5)/2]
log(y) = (5 - √5)/2
Ln(y)/Ln(10) = (5 - √5)/2
Ln(y) = [(5 - √5)/2].Ln(10)
Ln(y) = Ln(10)^[(5 - √5)/2]
→ y = 10^[(5 - √5)/2]
Second case: log(x) = (5 - √5)/2 and log(y) = (5 + √5)/2
log(x) = (5 - √5)/2
Ln(x)/Ln(10) = (5 - √5)/2
Ln(x) = [(5 - √5)/2].Ln(10)
Ln(x) = Ln(10)^[(5 - √5)/2]
→ x = 10^[(5 - √5)/2]
log(y) = (5 + √5)/2
Ln(y)/Ln(10) = (5 + √5)/2
Ln(y) = [(5 + √5)/2].Ln(10)
Ln(y) = Ln(10)^[(5 + √5)/2]
→ y = 10^[(5 + √5)/2]

[@RealQinnMalloryu4]

(y ➖ 3x+2) (y ➖ 5x+1).

[@TSR1942]

Hello, gentleman use a better pen . I want letters to brighter.

[@superacademy247]

Okay. I've started using thicker pen