Question 289496
Hi there. I need to obtain y in terms of x for the following logarithmic equation: 
ln(y-1) = 2 ln x + ln y-x 
The book answer (given without a working) is y = 1 / (1 - x^2 * E^-x),
but using the logarithmic identities I can only seem to get as far as
y = 1 + x^2 + yE^-x. I'm not sure if my error is in the log identities or the basic algebra. Thanks!

ok, let me try

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

ln is the natural log, it is base e, where e is Euler's Number approx.= 2.7183
base b = e (by definition)
ln(mn) = ln(m) + ln(n) (logarithmic rule)
ln(m/n) = ln(m) - ln(n) (logarithmic rule)
ln(n) = m --> e^m = n (logarithmic rule)
ln(m^n) = n * ln(m) (logarithmic rule)
e^(ln(m)) = m (logarithmic rule)
ln(e^m) = m (logarithmic rule)

ln(y - 1) = ln(x^2) + ln(y - x)
ln(y - 1) = ln(x^2 * (y - x))
ln(y - 1) = ln(x^2 * y - x^3)
you can set y - 1 = yx^2 - x^3
y - yx^2 = 1 - x^3
y * (1 - x^2) = 1 - x^3
y = (1 - x^3)/(1 - x^2) this does not appear to be the book answer you gave earlier,...

... so also evaluating ln(y - 1) = 2ln(x) + ln(y) - x
ln(y - 1) = ln(x^2) + ln(y) - x
ln(y - 1) = ln(yx^2) - ln(e^x)
ln(y - 1) = ln[(yx^2)/(e^x)]
you can set y - 1 = (yx^2)/(e^x)
(y - 1)/(yx^2) = 1/e^x = e^(-x)
y/(yx^2) - 1/(yx^2) = e^(-x)
y/y - 1/y = x^2 * e^(-x)
1 - 1/y = x^2 * e^(-x)
-1/y = x^2 * e^(-x) - 1
1/y = 1 - x^2 * e^(-x)
y = 1/[1 - x^2 * e^(-x)]
oh that gives that book answer you gave earlier