SOLUTION: I am really stuck with this. I need to find x in terms of y by solving{{{y=2e^(x+1)}}} I have come up with{{{x=ln(y)-ln(2e)}}}which does not look right also{{{x=((1/2)y-1)/ln(e)

Algebra ->  Algebra  -> Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I am really stuck with this. I need to find x in terms of y by solving{{{y=2e^(x+1)}}} I have come up with{{{x=ln(y)-ln(2e)}}}which does not look right also{{{x=((1/2)y-1)/ln(e)      Log On

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Question 20134: I am really stuck with this.
I need to find x in terms of y by solvingy=2e%5E%28x%2B1%29
I have come up withx=ln%28y%29-ln%282e%29which does not look right
alsox=%28%281%2F2%29y-1%29%2Fln%28e%29.what am I doing wrong?
Thanks in advance for any help.
Answer by AnlytcPhil(1116) About Me  (Show Source):
You can put this solution on YOUR website!


y=2ex+1

Divide both sides by 2

y/2 = ex+1

Use the rule of logs which says A = eB can be written as B = ln(A)

x+1 = ln(y/2)

x = ln(y/2) - 1

Edwin