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Mathematics 8 Online
OpenStudy (anonymous):

Solve ln x − ln(x −1)= 1

sam (.sam.):

ln x − ln(x −1)= 1 \[\ln(\frac{x}{x-1})= 1\] \[\frac{x}{x-1}=e^1\] x=ex-e x-ex=-e x(1-e)=-e \[x=\frac{-e}{1-e}\]

OpenStudy (anonymous):

what formula did you use?

sam (.sam.):

From where to where?

sam (.sam.):

ln x − ln(x −1)= 1 \[\ln(\frac{x}{x-1})= 1\] Here?

OpenStudy (anonymous):

the first step

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

e^a=b?

sam (.sam.):

See the attachment

OpenStudy (anonymous):

oh ok thanx

sam (.sam.):

ln x − ln(x −1)= 1 Using rule , log(m/n) = log(m) – log(n) \[\ln(\frac{x}{x-1})= 1\] Exponent both sides, \[e^{\ln(\frac{x}{x-1})}=e ^{1}\] \[\frac{x}{x-1}=e^1\] x=ex-e x-ex=-e x(1-e)=-e \[x=\frac{-e}{1-e}\]

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