Find $\log _{ 10 }{ { n }^{ m } }$

Thanks for the Information ,I did not knew such a possibility existed .

Let $a = m + 1$ and $b = n + 1$ . Then we have that

$\dfrac{1}{a} - \dfrac{1}{b} = \dfrac{4}{77} \Longrightarrow 77b - 77a = 4ab \Longrightarrow$

$4ab + 77a - 77b = (2a - \frac{77}{2})(2b + \frac{77}{2}) + (\frac{77}{2})^{2} = 0 \Longrightarrow$

$(77 - 4a)(4b + 77) = 77^{2} = 7^{2} \times 11^{2}$ .

Then since $a,b \gt 0$ we need to distribute the factors of $77^{2}$ to $77 - 4a$ and $4b + 77$ in such a way that we get positive integer solutions for $a,b$ . For example,

$77 - 4a = 7^{2}, 4b + 77 = 11^{2} \Longrightarrow (a,b) = (7,11), (m,n) = (6,10)$ ,

$77 - 4a = 1, 4b + 77 = 77^{2} \Longrightarrow (a,b) = (19,1463), (m,n) = (18,1462)$ .

These turn out to be the only factor-splitting options that yield integer results, thus ruling out $7, 10, 11$ as solutions to this problem.

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