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Solve ln x − ln(x −1)= 1
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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 \[x=\frac{-e}{1-e}\]
what formula did you use?
From where to where?
ln x − ln(x −1)= 1 \[\ln(\frac{x}{x-1})= 1\] Here?
the first step
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yes
e^a=b?
See the attachment
oh ok thanx
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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