how do you solve this equation for x? ln x + ln (1+x) = 1?

Answer 1

Brute force using the formula in an Excel spreadsheet.

x= 1.2229

Plug this in

ln(1.2229) + ln (1.1.2229) = 1.000038

Answer 2

First, consolidate the left-hand terms, using the log relation

ln(a•b) = ln(a) + ln(b)

ln(x) + ln(1+x) = 1
ln[x(1+x)] = 1

Now exponentiate both sides, since exponentiation is the inverse of the log function.

e^ln[x(1+x)] = e^1
x(1+x) = e

Solve for x, using quadratic principles - complete the square.

x + x² = e
x² + 2•Â½•x + ½² = e + ½²
(x + ½)² = e + ½²
√(x + ½)² = √(e + ¼)
|x + ½| = √(e + ¼)
x + ½ = ±√(e + ¼)
x = ‾½ ± √(e + ¼)

Answer 3

Raise the whole equation by e.

exp[ ln x + ln (1+x) ] = e
x (1+x) = e
x^2 + x - e = 0

Then apply the quadratic formula.

Answer 4

ln [ (x) (x + 1) ] = 1
(x)(x + 1) = e^1
x² + x = 2.72
x² + x - 2.72 = 0
x = [- 1 ± √(1 + 10.88) ] / 2
x = [- 1 ± √(11.88) ] / 2
x = 1.22 , x = - 2.22

Answer 5

i havent done this in a while but this might be it:

X=((1/In) -1)/2