Question

2*lnx-lny=ln[x+(3/4)y] atunci x/y=? macar o idee va rog

Answer 1

2lnx-lny = ln[x+(3/4)y]

ln(x²) - lny = ln[x+(3/4)y]

ln(x²/y) = ln[x+(3/4)y]

x²/y = x+(3/4)y |:x

x/y = 1+(3/4x)y

x/y = 1+(3/4)•(y/x)

Notam x/y = t => y/x = 1/t

t = 1+(3/4)•(1/t) |•4

4t = 4+3•(1/t) |•t

4t² = 4t+3

4t²- 4t - 3 = 0

Delta = 16 + 48 = 64

t₁,₂ = (4±8)/8, dar t > 0.

=> t = 12/8 = 3/2 => x/y = 3/2