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

Sep 16, 2016

#### Answer:

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

#### Explanation:

To keep logarithms as real numbers, $x > 0 \mathmr{and} y > 1$.

Rearranging,

$\ln \left(y - 1\right) - 3 \ln x = \ln \left(y - 1\right) - \ln \left({x}^{3}\right) = \ln \left(\frac{y - 1}{x} ^ 3\right) = 2$

Inversely,

$\frac{y - 1}{x} ^ 3 = {e}^{2}$. cross multiplying and rearranging,

$y = {e}^{2} {x}^{3} + 1$

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 $\left(- \frac{1}{e} ^ \left(\frac{2}{3}\right) , 0\right) .$