solve for x and y: jskskkaka
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(32)^x/2 = 2^y + 1
(2)^5×x/2
2^5x/2 = 2^y +1
Bases are same so equate the power
5x / 2 = y + 1
5x = 2y + 2
5x - 2y = 2. eq1
Now for 2nd equation
16^(8-x) / 2 = 8^y
2^4(8 - x)/2 = 2^3y
Now bases are same so equate the power
2(8 - x) = 3y
16 - 2x = 3y
2x + 3y = 16. eq 2
5x - 2y = 2. eq 1
so multiply eq 1 by 3 and eq2 by 2
15x - 6y = 6
4x + 6y = 32
19x = 38
x = 38 / 19 = 2
Put x in eq1
5 × 2 - 2y = 2
- 2y = 2 - 10
y = -8 / -2
y = 4
So x = 2 and y = 4
Thanks
(2)^5×x/2
2^5x/2 = 2^y +1
Bases are same so equate the power
5x / 2 = y + 1
5x = 2y + 2
5x - 2y = 2. eq1
Now for 2nd equation
16^(8-x) / 2 = 8^y
2^4(8 - x)/2 = 2^3y
Now bases are same so equate the power
2(8 - x) = 3y
16 - 2x = 3y
2x + 3y = 16. eq 2
5x - 2y = 2. eq 1
so multiply eq 1 by 3 and eq2 by 2
15x - 6y = 6
4x + 6y = 32
19x = 38
x = 38 / 19 = 2
Put x in eq1
5 × 2 - 2y = 2
- 2y = 2 - 10
y = -8 / -2
y = 4
So x = 2 and y = 4
Thanks
ashishtiwari14:
thanks bhai
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