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Robert Israel
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If $\log_x(y) = a/b$$\log_x(y) = j/k$ is rational, this reduces to a polynomial in $x^{1/b} = y^{1/a}$$x^{1/k} = y^{1/j}$. Otherwise you're unlikely to get a closed form. You might use numerical methods, or a series expansion: if $y = x^r$, $$ n = \frac{\ln(-c/a)}{\ln(x)} + \sum_{k=0}^\infty \frac{(-c/a)^{kr}(b/c)^k}{k! \ln(x)} \prod_{j=1}^{k-1} (kr - j)$$

If $\log_x(y) = a/b$ is rational, this reduces to a polynomial in $x^{1/b} = y^{1/a}$. Otherwise you're unlikely to get a closed form. You might use numerical methods, or a series expansion: if $y = x^r$, $$ n = \frac{\ln(-c/a)}{\ln(x)} + \sum_{k=0}^\infty \frac{(-c/a)^{kr}(b/c)^k}{k! \ln(x)} \prod_{j=1}^{k-1} (kr - j)$$

If $\log_x(y) = j/k$ is rational, this reduces to a polynomial in $x^{1/k} = y^{1/j}$. Otherwise you're unlikely to get a closed form. You might use numerical methods, or a series expansion: if $y = x^r$, $$ n = \frac{\ln(-c/a)}{\ln(x)} + \sum_{k=0}^\infty \frac{(-c/a)^{kr}(b/c)^k}{k! \ln(x)} \prod_{j=1}^{k-1} (kr - j)$$

Source Link
Robert Israel
  • 54.2k
  • 1
  • 76
  • 152

If $\log_x(y) = a/b$ is rational, this reduces to a polynomial in $x^{1/b} = y^{1/a}$. Otherwise you're unlikely to get a closed form. You might use numerical methods, or a series expansion: if $y = x^r$, $$ n = \frac{\ln(-c/a)}{\ln(x)} + \sum_{k=0}^\infty \frac{(-c/a)^{kr}(b/c)^k}{k! \ln(x)} \prod_{j=1}^{k-1} (kr - j)$$