lnx+ln(x+1)=1

Question

solving for x 

Answer 1

ln(x) + ln(x+1) = 1

 

x > 0

 

Use Multiplication Rule of Logs

 

ln(x(x+1)) = 1

 

Change to equivalent exponential form

 

e^1 = x(x+1) = x^2 + x

 

x^2 + x - e = 0

 

x = (-1 ± √(1^2-4*1(-e) ) )/(2*1)

 

x = (-1 ± √(1+4e) )/2

 

Since x > 0,

 

x = (-1 + √(1+4e) )/2 ≈ 1.22287022972105

 

check:

 

ln(1.22287022972105) + ln(1.22287022972105 + 1) ≈? 1

 

ln(1.22287022972105) + ln(1.22287022972105 + 1) ≈? 1

 

0.201200742915763 + 0.798799257084243 ≈? 1

 

1.00000000000001 ≈ 1 √

Answer 2

Hi Tatianna;

In x + In (x+1)=1

1=In_ee

In x + In (x+1)=In e

Let's apply the principle...

In ab= In a + In b

In x(x+1)=In e

Let's remove the Natural Logarithms...

x(x+1)=e

x²+x=e

Steve is correct.  I made a mistake.