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Mathematics 14 Online
OpenStudy (sphott51):

solve e^2x-1 = 3 for x

OpenStudy (johnweldon1993):

Is this \[\large e^{2x - 1} = 3\] or \[\large e^{2x} - 1 = 3\]

OpenStudy (sphott51):

e^(2x-1) = 3

OpenStudy (johnweldon1993):

Alright \[\large e^{2x - 1} = 3\] Remember the rules for logarithms...\(\large ln(e^{ax}) = ax \times ln(e)\) So if we apply this here, and take the log of both sides...we have \[\large ln(e^{2x - 1}) = ln(3)\] which then turns into \[\large (2x - 1)ln(e) = ln(3)\] Still with me?

OpenStudy (sphott51):

I am

OpenStudy (johnweldon1993):

Great...what is \(\large ln(e) = ?\)

OpenStudy (sphott51):

1?

OpenStudy (johnweldon1993):

Right! So we really just have \[\large (2x - 1) = ln(3)\] Now just solve that for 'x'

OpenStudy (sphott51):

Ln3 +1 / 2

OpenStudy (johnweldon1993):

Right! *just to make sure you mean all over 2 though...

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