Natural log x and y problem

A180S

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Please help solving this natural log problem.

Real variable x and y are repeated by the equation

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

A) Determine the range of values of x and y tor which the expression on each side of the equation is defined.

B) Find y explicitly as a function of x, that is, express the equation in the form y=f(x)
 
What bit are you stuck on?

Have you thought about what values of x would make the RHS undefined? or values of y that would make the LHS undefined?

Have you tried using some of the log laws to rewrite the expressions on either side of the equality?
 
*typo should be related not repeated.
Tbh I was given a paper from someone who did advanced maths and have been determined to work it out during lockdown, would like to see the inner workings of a solution to question a and b
 
Please help solving this natural log problem.
Real variable x and y are repeated by the equation
3ln(y-1) = ln(x-2) + In(x+2) - ln(x+1)
A) Determine the range of values of x and y tor which the expression on each side of the equation is defined.
B) Find y explicitly as a function of x, that is, express the equation in the form y=f(x)
A) 3log(y1)=log(y1)33\log(y-1)=\log(y-1)^3 because logarithm function is defined on on positive numbers y>1y>1
So because we have log(x2)\log(x-2) we must x>2x>2.

B) Writing in compact form we get.
3log(y1)=log(x2+)log(x+2)log(x+1)log(y1)3=log((x2)(x+2)(x+1))log(y1)3=log((x24)(x+1))3\log(y-1)=\log(x-2+)\log(x+2)-\log(x+1)\\\log(y-1)^3=\log\left(\dfrac{(x-2)(x+2)}{(x+1)}\right)\\\log(y-1)^3=\log\left(\dfrac{(x^2-4)}{(x+1)}\right)
Because the natural logarithm, log(x)\log(x), is one-to-one we get:
(y1)3=(x24)(x+1)(y-1)^3=\dfrac{(x^2-4)}{(x+1)} OR y=x24x+13+1\Large{y=\sqrt[3]{\dfrac{x^2-4}{x+1}}+1}
 
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