SOLUTION: If e^(2x)-9e^(x)+18=0 , then the solutions (arranged in increasing order of x) are x=_______and x=________

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Question 115289: If e^(2x)-9e^(x)+18=0 , then the solutions (arranged in increasing order of x) are x=_______and x=________

Answer by ganesh(20) About Me  (Show Source):
You can put this solution on YOUR website!
Our equation is e^2x - 9e^x + 18 = 0.
That is, (e^x)^2 - 9e^x + 18 = 0 -------(1)
Let y = e^x.
So, (1) becomes, y^2 - 9y + 18 = 0.
That is, (y- 3) (y -6) = 0.
Or, y = 3 and y = 6.
Or, e^x = 3 and e^x = 6.
That is, Log e^x = Log3 and Log e^x = Log6.
That is, x = Log3 and x = Log6