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Mathematics 22 Online
OpenStudy (anonymous):

Solve for x. ln(2x-1)=0

OpenStudy (anonymous):

x=1

OpenStudy (anonymous):

one

OpenStudy (anonymous):

how did you get that answer?

OpenStudy (anonymous):

x=0 or x=1

OpenStudy (anonymous):

zero or one

OpenStudy (anonymous):

first x=0 and secondaly 2nd term is also zero after exponential raise to zero is also one then u calculate

OpenStudy (across):

Exponentiate both sides: \[2x-1=1\]\[2x=2\]\[x=1\]

OpenStudy (anonymous):

Thanks

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