Question (2 Viewers)

[aqwerty13402]

Can someone pls answeer this im so lost
Given that ln(a) - ln(b) = ln(a-b), where a > b > 1, show that a = b^2/(b-1)

 

[liamkk112]

Can someone pls answeer this im so lost
Given that ln(a) - ln(b) = ln(a-b), where a > b > 1, show that a = b^2/(b-1)

ln(a/b) =ln(a-b) by log laws
so a/b = a-b
a = ab - b^2
b^2 = ab-a = a(b-1)
so a = b^2/(b-1)

 

[aqwerty13402]

ln(a/b) =ln(a-b) by log laws
so a/b = a-b
a = ab - b^2
b^2 = ab-a = a(b-1)
so a = b^2/(b-1)

THANKSSS