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

Question

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

1 Answer

Impact of this question

8327 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-16T14:45:48

$x = \frac{\sqrt{1 + 4 e} - 1}{2} = 1.223$ $x=(sqrt(1+4e)-1)/2=1.223$

Explanation:

$ln x + ln (x + 1) = 1$ $lnx+ln(x+1)=1$

Hence $x (x + 1) = e$ $x(x+1)=e$

i.e. $x^{2} + x - e = 0$ $x^2+x-e=0$

and using quadratic formula

$x = \frac{- 1 \pm \sqrt{1^{2} + 4 e}}{2}$ $x=(-1+-sqrt(1^2+4e))/2$

$= \frac{- 1 \pm \sqrt{1 + 4 e}}{2}$ $=(-1+-sqrt(1+4e))/2$

$= \frac{- 1 \pm 3.446}{2}$ $=(-1+-3.446)/2$

$= \frac{2.446}{2}$ $=(2.446)/2$ or $- \frac{4.446}{2}$ $-4.446/2$

$= 1.223$ $=1.223$ or $- 2.223$ $-2.223$

But as $x = - 2.223$ $x=-2.223$ is not in domain, we have $x = \frac{\sqrt{1 + 4 e} - 1}{2} = 1.223$ $x=(sqrt(1+4e)-1)/2=1.223$
graph{lnx+ln(x+1)-1 [-5, 5, -2.5, 2.5]}

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

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