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 $x$.

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

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

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

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

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

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

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