I Solve for ##n## in ##\frac{1}{(T+\frac{1}{U^{1/n}})^n} = G##

[MevsEinstein]

TL;DR

What the title says

I was doing some research on Space where I stumbled on this equation: $\frac{1}{(T+s)^n} = G$. $T$ is independent and $G$ is dependent. $s$ and $n$ are constants. I found out what $s$ was ($\frac{1}{U^{1/n}}$), and so I substituted it into the equation. Now, I need to find $n$ in terms of other variables that aren't $n$. How could I do this?

 

[Mark44]

Summary: What the title says

I was doing some research on Space where I stumbled on this equation: $\frac{1}{(T+s)^n} = G$. $T$ is independent and $G$ is dependent. $s$ and $n$ are constants. I found out what $s$ was ($\frac{1}{U^{1/n}}$), and so I substituted it into the equation. Now, I need to find $n$ in terms of other variables that aren't $n$. How could I do this?

Your equation is equivalent to $(T + s)^n = \frac 1 G$, assuming that $G \ne 0$. Now take the log (in whatever base) of both sides to isolate n.

 

[mathman]

nln(T+s)=-ln(G) or n=-ln(G)/ln(T+s).

 

[Mark44]

nln(T+s)=-ln(G) or n=-ln(G)/ln(T+s).

Pretty much what I said, except that I was going to let the OP do a little of the work.

 

[Delta2]

s depends on n too, it is $s=\frac{1}{U^{\frac{1}{n}}}$ which seems to complicate things.

 

[mfb]

That won't have a nice exact solution but a numerical approach should work well.
Maybe there is some useful approximation if we know more about the values.