What are the steps to solving a system of equations when $x$ and $y$ are exponents? But they have different base. Here is the problem.



  • 2
    $\begingroup$ Multiply the two euqtions: You get a value for $x+y$. Divide the two equations, you get a value for $x-y$ $\endgroup$
    – voldemort
    Jan 22, 2015 at 1:09
  • $\begingroup$ What should I multiply with to get the value of $x+y$? $\endgroup$ Jan 22, 2015 at 5:58
  • $\begingroup$ Multiply the first equation to the second equation. $\endgroup$
    – velut luna
    Jan 22, 2015 at 6:11
  • $\begingroup$ Super thank you voldemort and @kyson, I found the solution. I got $x=1$ and $y=2$. $\endgroup$ Jan 23, 2015 at 1:29

1 Answer 1


Take log.



  • $\begingroup$ I'm sorry, I'm still confuse how to find the value of $x$ and $y$. Since the $x$ and $y$ are the power for different number, I can't manipulate them. And after taking logarithm, though, it couldn't help me yet. $\endgroup$ Jan 22, 2015 at 4:12
  • $\begingroup$ Do you know how to solve simultaneous linear equations? Otherwise, you can follow voldmort's method. $\endgroup$
    – velut luna
    Jan 22, 2015 at 5:40
  • $\begingroup$ I know how to solve simultaneous linear equations. But I don't know how to modify these equations as they have logarithm in it. And, what is voldmort's method? $\endgroup$ Jan 22, 2015 at 5:55
  • $\begingroup$ The method suggested by voldmort in his comment on your post. $\endgroup$
    – velut luna
    Jan 22, 2015 at 6:04
  • $\begingroup$ If you know how to simultaneous linear equations, then you should be able to solve the equations in the answer I posted. It's just two simultaneous linear equations. The coefficients are logarithms though. But logarithms are just numbers. $\endgroup$
    – velut luna
    Jan 22, 2015 at 6:06

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