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Solve for $x,n$
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candelo6a
$\frac{b}{a}={\left(\frac{x+1}{x}\right)}^{n}$
${\left(\frac{b}{a}\right)}^{\frac{1}{n}}=1+\frac{1}{x}$
$x=\frac{1}{{\left(\frac{b}{a}\right)}^{\frac{1}{n}}-1}$
Substitute this in $\left(1\right)$ to find $n$.
Now use since $x$ is an integer and given $x=\frac{1}{{\left(\frac{b}{a}\right)}^{\frac{1}{n}}-1}$, what can you infer?
EDIT: As x is a positive integer as is said by OP in the comments, hence ${\left(\frac{b}{a}\right)}^{\frac{1}{n}}-1=1$, or $b=a{2}^{n}$ or $n=\frac{log\frac{b}{a}}{log2}$.
$\therefore n=\frac{log\frac{b}{a}}{log2}$ and $x=1$