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Mathematics 7 Online
OpenStudy (anonymous):

2lnx-1/2/e^x=0

OpenStudy (anonymous):

\[2 \ln (x)-\frac{e^x}{2}=0\] Is this right?

OpenStudy (anonymous):

unsolvable

OpenStudy (anonymous):

\[2lnx-1\]/\[2-e^x\]

OpenStudy (anonymous):

i can give you an iteration,

myininaya (myininaya):

is this what you have written? \[\frac{2lnx-1}{2-e^x}=0\] \[2lnx-1=0\] \[2lnx=1\] \[lnx=\frac{1}{2}\] \[x=e^\frac{1}{2}\]

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