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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} $

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  • $\begingroup$ Subtracting the equations gives $y^{70} - y^{50} = 499,$ which is a trinomial equation. For a rough idea of the nature of the solutions, you can consider rough hand-sketches of $y = x^{70}$ and $y = x^{50} + 499,$ and observe that by general principles of what the graphs look like there will be two points where the graphs intersect, one for a negative value of $x$ (hence, a negative value of $y$ in your equations) and the other for a positive value of $x$ (hence, a positive value of $y$ in your equations). $\endgroup$ – Dave L. Renfro Jul 2 at 21:23
  • $\begingroup$ Also, this is an algebraic system of equations, not an exponential system. $\endgroup$ – Dave L. Renfro Jul 2 at 21:25
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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.

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  • $\begingroup$ Thanks. I used this to find u = 2.49085 so x = 95.88217 and y = 1.095556. If I take this one step further and add one more equation which is y^80 = 800+x, how should I approach that? $\endgroup$ – memokerobi Jul 3 at 20:04
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    $\begingroup$ then you would have three equations in the two variables $x,y$, which in the general case cannot be simultaneously verified. $\endgroup$ – G Cab Jul 3 at 20:14
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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.

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