Natural log x and y problem

[A180S]

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)

 

[Harry_the_cat]

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?

 

[A180S]

*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

 

[pka]

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) \(3\log(y-1)=\log(y-1)^3\) because logarithm function is defined on on positive numbers \(y>1\)
So because we have \(\log(x-2)\) we must \(x>2\).

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