How do you solve for y in #Ln (y-1) = 3 ln x +2#?

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Answer 1 · 2016-09-16T07:05:24

With #x > 0 and y >1, y=e^2x^3+1#.
The graph is that part in Q1, above (0, 1), sans (0, 1)..

To keep logarithms as real numbers, #x > 0 and y > 1#.

Rearranging,

#ln(y-1)-3ln x= ln(y-1)-ln(x^3)=ln((y-1)/x^3)=2#

Inversely,

#(y-1)/x^3=e^2#. cross multiplying and rearranging,

#y=e^2x^3+1#

The graph is the part of this curve in #Q_1

and is above (0, 1), sans (0, 1).

Of course, the whole curve ( not the given part ) meets the x-axis y

= 0, at #(-1/e^(2/3), 0).#