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

sunshine022uv

sunshine022uv

Answered question

2021-12-14

How do you solve lnx+ln(x1)=1 ?

Answer & Explanation

MoxboasteBots5h

MoxboasteBots5h

Beginner2021-12-15Added 35 answers

Making use of the logarithms' rules,
ln(x)+ln(x1)=ln(x(x1))=ln(x2x) 
Therefore, 
ln(x2x)=1 
Then, both sides are exponentiated (raised to the e power) as follows:
eln(x2x)=e1 
Simplify while keeping in mind that exponents reverse logarithms:
x2x=e 
Now, we complete the square: 
x2x+14=e+14 
Simplify: 
(x12)2=e+14=4e+14 
Add the square roots of both sides together:
x12=±4e+12 
Add 12 to both sides: 
x=1±4e+12 
Eliminate the negative answer (in logab,b>0): 
x=1+4e+12

Joseph Lewis

Joseph Lewis

Beginner2021-12-16Added 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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