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

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

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

The solution is x=sqrt(e-1),x=-sqrt(e-1)

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

x = e - 1, x = - e - 1

Decimal Notation

x = 1.31083 \dots, x = - 1.31083 \dots

Solution steps

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

Apply log rules

1 + x^{2} = e

Solve

1 + x^{2} = e : x = e - 1, x = - e - 1

x = e - 1, x = - e - 1

10 = 20 lo g_{10} (\frac{2}{x})

y, ln (y (k (x + 1))) = x

- x - 2 = ln (x - 1)

x, ln (x - y) = 0.5

ln (\frac{1.5}{x}) = 27

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