An equation is given in this mathematics problem and it contains the literals $a$, $n$ and $x$ but the equation expresses the value of $x$ in terms of a raised power of $n$ and $–n$ in sum, difference and division form.

$x = \dfrac{a^n -a^{-n}}{a^n+a^{-n}}$

This equation has be transformed as another equation and it should represent the value of $n$ in terms of $x$. This mathematical problem can be solved in two different methods.

Use cross multiplication method to start simplifying this equation.

$\implies x(a^n+a^{-n}) = a^n -a^{-n}$

$\implies x.a^n+x.a^{-n} = a^n -a^{-n}$

$\implies x.a^{-n}+a^{-n} = a^n -x.a^n$

$\implies (x+1)a^{-n} = (1-x)a^n$

$\implies (1+x)a^{-n} = (1-x)a^n$

Express this equation in fraction form but the $x$ terms should be one side and $n$ terms at other side.

$\implies \dfrac{(1+x)}{(1-x)} = \dfrac{a^n}{a^{-n}}$

Now use division rule of exponents to simplify the right hand side expression.

$\implies \dfrac{(1+x)}{(1-x)} = a^{(n-(-n))}$

$\implies \dfrac{1+x}{1-x} = a^{(n+n)}$

$\implies \dfrac{1+x}{1-x} = a^{2n}$

$\implies a^{2n} = \Bigg[\dfrac{1+x}{1-x}\Bigg]$

Apply the fundamental relation of logarithm with exponential form and express exponential form expression in terms of the logarithm.

$\implies 2n = \log_{a} \Bigg[\dfrac{1+x}{1-x}\Bigg]$

$\therefore \,\,\,\,\,\, n = \dfrac{1}{2} \, \log_{a} \Bigg[\dfrac{1+x}{1-x}\Bigg]$

It is the required solution for this mathematics problem.

The problem can also be solved in another method alternatively.

$x = \dfrac{a^n -a^{-n}}{a^n+a^{-n}}$

Express the equation in reciprocal form.

$\implies \dfrac{1}{x} = \dfrac{a^n+a^{-n}}{a^n -a^{-n}}$

Use componendo and dividendo rule to simply this equation.

$\implies \dfrac{1+x}{1-x} = \dfrac{a^n+a^{-n}+a^n -a^{-n}}{a^n+a^{-n} -(a^n -a^{-n})}$

$\require{cancel} \implies \dfrac{1+x}{1-x} = \dfrac{a^n+\cancel{a^{-n}}+a^n -\cancel{a^{-n}}}{\cancel{a^n}+a^{-n} -\cancel{a^n} +a^{-n}}$

$\implies \dfrac{1+x}{1-x} = \dfrac{2a^n}{2a^{-n}}$

$\require{cancel} \implies \dfrac{1+x}{1-x} = \dfrac{\cancel{2}a^n}{\cancel{2}a^{-n}}$

$\implies \dfrac{1+x}{1-x} = \dfrac{a^n}{a^{-n}}$

Now, repeat the previous method’s procedure to get the value of $n$ in terms of $x$.

$\implies \dfrac{(1+x)}{(1-x)} = a^{(n-(-n))}$

$\implies \dfrac{1+x}{1-x} = a^{(n+n)}$

$\implies \dfrac{1+x}{1-x} = a^{2n}$

$\implies a^{2n} = \Bigg[\dfrac{1+x}{1-x}\Bigg]$

$\implies 2n = \log_{a} \Bigg[\dfrac{1+x}{1-x}\Bigg]$

$\therefore \,\,\,\,\,\, n = \dfrac{1}{2} \, \log_{a} \Bigg[\dfrac{1+x}{1-x}\Bigg]$

Latest Math Topics

Mar 21, 2023

Feb 25, 2023

Feb 17, 2023

Feb 10, 2023

Latest Math Problems

Mar 03, 2023

Mar 01, 2023

Feb 27, 2023

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the math problems in different methods with understandable steps and worksheets on every concept for your practice.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved