^2log(xy)=7 dan ^2log(x^2/y)=5, maka x + y? pakai cara

[tex]^2\log{xy}=7\\^2\log{x}+^2\log{y}=7\\^2\log{x}=7-^2\log{y}\dots (1)[/tex]

[tex]^2\log{ \frac{x^2}{y} }=5\\^2\log{x}^2-^2\log{y}=5\\2\times^2\log{x}-^2\log{y}=5\\2\times(7-^2\log{y})-^2\log{y}=5\\14-3\times^2\log{y}=5\\3\times^2\log{y}=9\\^2\log{y}=3\dots (2)\\ \\ \frac{\log{y}}{\log{2}} = 3\\\log{y}=3\times\log{2}\\ \log{y}=\log{2^3}=\log{8}\\y=8[/tex]

Dari persamaan (1):
[tex]^2\log{x}=7-^2\log{y}[/tex]
Masukkan persamaan (2):
[tex]^2\log{x}=7-3 = 4\\ \frac{\log{x}}{\log{2}} =4\\\log{x}=4\times\log{2}=\log{16}\\x=16[/tex]
Maka,
[tex]x+y=16+8=24[/tex]