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Gomathi
Answers (2)
9^x + 3^y =10. [x=y-1] 3^2x + 3^y =10 3^(2y-2) + 3^y =10 3^2y / 3^2 + 3^y = 10 3^2y + (3^2)(3^y) =10 Here assume 3^y=u u² + 9u -10=0 Using quadratic function u=6 or -15 (Skip -ve value ) u=6. 3^y=6 log base 3 of 6= y Change base value lg6/lg3 =y y=1.63
I didn't understand how did you get (3^y+15) (3^y-6)
