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Question Number 63915 by gunawan last updated on 11/Jul/19

$$\mathrm{If}p\mathrm{is}\mathrm{nearly}\mathrm{equal}\mathrm{to}q\mathrm{and}n>1,\mathrm{such}$$ $$\mathrm{that}\frac{(n+1)p+(n-1)q}{(n-1)p+(n+1)q}={\left(\frac{p}{q}\right)}^k,\mathrm{then}\mathrm{the}$$ $$\mathrm{value}\mathrm{of}k\mathrm{is}$$

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