How to solve this multivariable exponential equation?

Question

I searched if this was asked before but couldn't find a solution. I have this equation

$y^{70} = x + 500 $

$y^{50} = x + 1 $

Is it possible to solve this equation? The only thing I could do is to bring it into this form and then cross-multiply which didn't yield many results.

$y^{20} = \frac{x+500}{x+1} $

Answer 1

The best I can think of at the moment is

$$ \eqalign{ & \left\{ \matrix{ 50\ln y = \ln \left( {1 + x} \right) \hfill \cr 20\ln y = \ln \left( {{{x + 500} \over {x + 1}}} \right) = \ln \left( {1 + {{499} \over {x + 1}}} \right) \hfill \cr} \right. \cr & \ln y = {1 \over {50}}\ln \left( {1 + x} \right) = {1 \over {20}}\ln \left( {1 + {{499} \over {x + 1}}} \right) \cr & \left( {1 + x} \right)^{\,2/5} = 1 + {{499} \over {x + 1}} \cr & \left( {1 + x} \right)^{\,7/5} = x + 1 + 499 \cr & 1 + x = u^{\,5} \cr & u^{\,7} - u^{\,5} = u^{\,5} \left( {u^{\,2} - 1} \right) = 499 \cr} $$

which clearly has only one solution, and that numerically is easy to solve.

Answer 2

Raise the first one to the fifth power to get $$y^{350}=(x+500)^5$$

Raise the second one to the seventh power to get $$ y^{350}=(x+1)^7$$

Thus you have $$(x+500)^5=(x+1)^7$$ This is a seventh degree polynomial to solve.