math

solve for x and y , the simultaneous equations 4^x+3 = 32(2^x+y) and 9^x + 3^y = 10

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1. Assuming the usual carelessness with parentheses, I get

4^(x+3) = 32(2^(x+y))
2^(2(x+3)) = 2^5*2^(x+y)
2x+6 = 5+x+y
x = y-1

9^(y-1) + 3^y = 10
3^(2y-2) + 3^y = 10
3^(2y)/9 + 3^y = 10
3^(2y) + 9*3^y - 90 = 0
(3^y + 15)(3^y - 6) = 0
3^y = 6
y = log6/log3
x = y-1

That answer is unusual, so check to make sure I got things right.

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y = 1.6309, x = 0.6309

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4. How did u get (3^y + 15)(3^y - 6) = 0 ?

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