What is the answer to ln(x^2+x)+1=ln(x+1) ?

The answer to ln(x^2+x)+1=ln(x+1) is x=(-e+1+sqrt(e^2+2e+1))/(2e)

Solve for x: ln(x^2+x)+1=ln(x+1)

The solution is x=(-e+1+sqrt(e^2+2e+1))/(2e)

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ln (x^{2} + x) + 1 = ln (x + 1)

x = \frac{- e + 1 + e ^{2} + 2 e + 1}{2 e}

Decimal Notation

x = 0.36787 \dots

Solution steps

ln (x^{2} + x) + 1 = ln (x + 1)

Apply log rules

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

Solve

(x^{2} + x) e = x + 1 : x = \frac{- e + 1 + e ^{2} + 2 e + 1}{2 e}, x = \frac{- e + 1 - e ^{2} + 2 e + 1}{2 e}

x = \frac{- e + 1 + e ^{2} + 2 e + 1}{2 e}, x = \frac{- e + 1 - e ^{2} + 2 e + 1}{2 e}

Verify Solutions:

x = \frac{- e + 1 + e ^{2} + 2 e + 1}{2 e}

True

, x = \frac{- e + 1 - e ^{2} + 2 e + 1}{2 e}

False

The solution is

x = \frac{- e + 1 + e ^{2} + 2 e + 1}{2 e}

ln (x) + ln (x - 1) = ln (4 - x)

ln (- 3 x + 5) = 6

lo g_{x} (\frac{1}{10}) = - 1

lo g_{4} (3 x - 5) = 3

lo g_{4} (3 x - 5) = 2

What is the answer to ln(x^2+x)+1=ln(x+1) ?
The answer to ln(x^2+x)+1=ln(x+1) is x=(-e+1+sqrt(e^2+2e+1))/(2e)
Solve for x: ln(x^2+x)+1=ln(x+1)
The solution is x=(-e+1+sqrt(e^2+2e+1))/(2e)