The equation is lnx+ln(x−1)=ln2 . I have worked it all the way through, and after factoring the x^2−1x−2 I got x=2, x=−1, but my question is: Can we have both solutions or couldn't we have the negatives?

Tyson Atkins

Tyson Atkins

Answered question

2022-10-25

Logarithmic Equations and solving for the variable
The equation is ln x + ln ( x 1 ) = ln 2 . I have worked it all the way through, and after factoring the x 2 1 x 2 I got x = 2, x = 1, but my question is: Can we have both solutions or couldn't we have the negatives?

Answer & Explanation

getrdone07tl

getrdone07tl

Beginner2022-10-26Added 23 answers

Using the property of logaritm ln a + ln b = ln ( a b ) we obtain
ln x ( x 1 ) = ln 2
Now, for the property ln u = ln v u = v the obtain that
x ( x 1 ) = 2 x 2 x 2 = 0
Then, we obtain x = 2 and x = 1, but x = 1 is not solution of the equation because ln ( 1 ) R (the function f ( x ) = ln x is defiined when x > 0).
Conclusion: the equation ln x + ln ( x 1 ) = ln 2 have the solution x = 2
Amiya Melendez

Amiya Melendez

Beginner2022-10-27Added 2 answers

The easiest way is indeed to combine them:
ln [ x ( x 1 ) / 2 ] = 0
implying that x ( x 1 ) = 2 or x 2 x 2 = 0, which indeed has exactly two solutions at x = 2 and x = 1
But x = 1 cannot be a solution, since the original equation is only defined for x 1, so the only solution is x = 2

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?