# Thread: Solving a difficult n-th root equation

1. ## Solving a difficult n-th root equation

How can I solve for n for something of this form?

(1.14)^(1/n) - 1 = 0.13/n

I managed to arrive at this:

n[(1.14)^(1/n) -1] = 0.13
But I don't know where to go from there.

2. Originally Posted by Volcanicrain
How can I solve for n for something of this form?

(1.14)^(1/n) - 1 = 0.13/n

I managed to arrive at this:

n[(1.14)^(1/n) -1] = 0.13
But I don't know where to go from there.
I don't know a way to solve this equation algebraically.

I would use Newton's method to get an approximate value for n:

$f(n)=n\left((1.14)^{\frac1n}-1\right)-0.13~,~n\in \mathbb{R}$

Solve $f(n)=0$ for n.

I got $n \approx -8.3045$