can u solve this simultaneous equation?

i) It is given lg; so let me take this as base 10 and not as e.

[For taking base e, generally it will be mentioned as ln]

ii) 2log(y) = log(y²) [Application of power law of log]

log(x) + 2log(y) = log(x) + log(y²) = log(xy²) [Application of product law of log]

So as given log₁₀(xy²) = 3

Hence, by definition of log, xy² = 10³ = 1000 ------ (1)

iii) Given x²y = 125 ---- (2)

iv) Dividing (1) by (2): y/x = 8; ==> y = 8x

Substituting in (2) above, 8x³ = 125

Taking cube root on both sides, 2x = 5

==> x = 5/2; so y = 8x = 8*(5/2) = 20

Thus x = 5/2 and y = 20

Logxy^2=3

xy^2=e^3-------(1) x^2y=125-------(2) so x^3y^3=125e^3 so xy=5e ----(3)

x^2y/xy=125/5e

x=25/e xy^2/xy=e^3/5e so y=e^2/5