How do you solve \ln x+\ln(x-1)=1 ?

sunshine022uv 2021-12-14 Answered
How do you solve lnx+ln(x1)=1 ?
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Expert Answer

MoxboasteBots5h
Answered 2021-12-15 Author has 35 answers
Using the rules of logarithms,
ln(x)+ln(x1)=ln(x(x1))=ln(x2x)
Therefore,
ln(x2x)=1
Then, we exponentiate both sides (put both sides to the e power):
eln(x2x)=e1
Simplify, remembering that exponents undo logarithms:
x2x=e
Now, we complete the square:
x2x+14=e+14
Simplify:
(x12)2=e+14=4e+14
Take the square root of both sides:
x12=±4e+12
Add 12 to both sides:
x=1±4e+12
Eliminate the negative answer (in logab,b>0):
x=1+4e+12

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Joseph Lewis
Answered 2021-12-16 Author has 43 answers
Simplify the left side.
ln(x2x)=1
To solve for x, rewrite the equation using properties of logarithms.
eln(x2x)=e1
Solve for x
x=1+1+4e2,11+4e2
Exclude the solutions that do not make ln(x)+ln(x1)=1 true.
x=1+1+4e2
The result can be shown in multiple forms.
Exact Form:
x=1+1+4e2
Decimal Form:
x=2.2228722

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