SO I assume you really meant this:
2^x+3^y=17
2^(x+2)+3^(y+1)=5
changing the second equation to
4*2^x +3*3^y = 5
now multiply the first equation by 4
4*2^x+4*3^y=17*4
now subtract
4(2^x) +3*3^y=5
and the result is
4*3^y-3*3^y=17*4-5
or
3^y=63=3^2 * 7
take the log(base3) of each side
Y=2 +log3 (7)
Plz help me to solve this system of equation.
2^x+3^y=17
2^x+2+3^y+1=5
2 answers
Represent 2 ^ x with p and 3 ^y with q . .p + q =17. And 2^ x+2 = 2^x X 2^2 =4p + 3 ^ y+1 =3^y x 3^1 =3q. . Solve simultaneously. P +q=17. . 4p + 3q = 5. . . . q =63 . P = -46
2 answers