ln (y-1) = 3 ln x +2. Solve in terms of y.?

Answer 1

using the properties:
e^(lnx) = x
and
ylnx = ln(x^y)
we can tear this bad-boy right to shreds:

ln(y-1) = 3lnx +2
exponentiate both sides:
e^(ln(y-1)) = e^(3lnx +2)
which is equivalent to:
y - 1 = [e^(3lnx)][e^2]
which reduces to:
y -1 = [e^(ln(x^3))](e^2)
which is equivalent to:
y-1 = (x^3)(e^2)
which is obviously:
y = (e^2)(x^3) + 1

Answer 2

ln (y-1) = 3 ln (x) +2

y-1 = e^(3 ln (x) +2)

y = 1+ x*e^(3) * e^(2)

Answer 3

ln(y-1)=3lnx+2=A
ln(y-1)=A , so we have y-1=e^A
3lnx+2=lnx^3 +2=A , then lnx^3=A-2
so we have x^3= e^(A-2) or x^3=e^A/e^2
then e^A=e^2 *x^3 , finally we have
y-1= e^2*x^3 or y=e^2*x^3+1

Answer 4

ln(y-1)=3lnx+2
ln(y-1)=lnx^3+lne^2
lnx^3=ln[(y-1)/e^2]
x^3=(y-1)/e^2
x=(y-1)^1/3/e^2/3.ANS