Jika x dan n>1 memenuhi log x = 2 logy dan ˣlogyˣ = ⁿlogxⁿ, maka x + n?

log x = 2 log y ==> log x = log y²

x = y²

ˣlogyˣ = (y)logx(y)

maaf gak ada aplikasi nulis lognya. subtitusikan x = y²

nanti didapatkan
y²/2 = 2y
y² = 4y
y = 4

x = y²
=4²
= 16

jadi x + y = 16 + 4 = 20

maaf kalau salah

log x = 2 log y
log x = log y^2
x = y^2

xlogy^x = ylogx^y
x . xlogy = y . ylogx
y^2 . y^2logy = y . ylogy^2
y^2 . 1/2 . ylogy = y . 2 . ylogy
1/2 y^2 = 2y
y^2 = 4y
y^2 - 4y = 0
y(y - 4) = 0
y = 0 atau y = 4
Karena syarat logaritma x > 0, y > 0 maka
y = 4
x = y^2 = 4^2 = 16
x + y = 16 + 4 = 20