Amazing. I messed up calculations trying some substitutions but intorducing a square was impressive. Still after substituting a and b i woukd not square just use the power of 1/2 as square root getting √x^√y=9=3^2 √y^√x=8=2^3 Wonder if this is sufficient to get the solutions or will itnkead to omission
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✓y=b , y=b^2 , ✓x=a , x=a^2 then
(a^2)^b-(b^2)^2=17 , (a^b-b^a)(a^b+b^a)=17 =1×17 , a^b-b^a=1 and a^b+b^a=17 , a^b=9=3^2 and b^a=8=2^3 , a=3 , b=2 and x=3^2=9 , y=2^2=4
Amazing. I messed up calculations trying some substitutions but intorducing a square was impressive.
Still after substituting a and b i woukd not square just use the power of 1/2 as square root getting
√x^√y=9=3^2
√y^√x=8=2^3
Wonder if this is sufficient to get the solutions or will itnkead to omission
th-cam.com/video/YrvZkHipU4o/w-d-xo.htmlsi=JjOLfow6SvhsVP4E