Solutions for logarithmic equation lnx+ln(x+1)=ln2
- print Print
- list Cite
Expert Answers
hala718
| Certified Educator
calendarEducator since 2008
write3,662 answers
starTop subjects are Math, Science, and Social Sciences
We know that lnx +lny = lnxy
==> lnx+ln(x+1) =ln2
==> lnx(x+1) = ln2
==> x(x+1)=2
==> x^2+x-2=0
==> (x+2)(x-1)=0
==> x=1
( We will not consider x=-2 as a solution, because the function is not defined for negative values.)
Related Questions
- Solutions for logarithmic equation ln(x+1)-ln(x-1)=ln2
- 3 Educator Answers
- Solve the logarithmic equation lnx - ln(x+1) = 2?
- 1 Educator Answer
- Simplify: i) ln(2x) + ln(2/x) ii) ln(x^2 - 1)- ln(x-1) Then solve: i) lnx = 5 ii) lnx + ln5 =...
- 2 Educator Answers
- #4. What is the derivative of y = x^(ln x) where ln x is the natural logarithm of x.
- 1 Educator Answer
- Which are solutions of equation lnx + ln(x+1)=2?
- 1 Educator Answer
neela | Student
To find the solutions to lnx+ln(x+1) = ln2.
Solution:
Since lna+lnb = ln(ab), we can write the equation as:
ln[x(x+1)] = ln2. Taking antilogarithm,
x(x+1) = 2.Or
x^2+x-2 = 0.
x^2+2x-(x+2) = 0
x(x+2)-1(x+2) = 0.
(x+2)(x-1) = 0. Or
x+2 = 0 . Or x-1 = 0.
x+2 = 0 gives: x = -2 not practical as ln is not defined for < 0 Or negative quantities.
x-1 = 0 gives: x= 1 which is the practical solution.
check Approved by eNotes Editorial
Student Answers