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 $(-1/e^(2/3), 0).$

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