SOLUTION: Solve for the unknowns simultaneously: 2logy = log2 + logx and 2^y = 4^x

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Question 390008: Solve for the unknowns simultaneously:
2logy = log2 + logx and 2^y = 4^x
Found 2 solutions by jim_thompson5910, scott8148:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Hint: Since 2%5Ey=4%5Ex, this means that 2%5Ey=%282%5E2%29%5Ex and 2%5Ey=2%5E%282x%29


So because 2%5Ey=2%5E%282x%29, this means that y=2x (ie since the bases are equal, the exponents are equal)


So the second equation becomes y=2x

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
2logy = log2 + logx ___ y^2 = 2x

2^y = 4^x ___ 2^y = (2^2)^x ___ y = 2x

substituting ___ y^2 = y ___ so y equals zero or one

the log of zero is undefined , so ___ y = 1

substituting ___ (1) = 2x ___ 1/2 = x