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

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

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

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

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

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

Decimal Notation

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

Solution steps

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

Apply log rules

x^{2} + 1 = e

Solve

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

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

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

ln (x^{2} + 1) = 3

lo g_{X} (81) = 4

3 lo g_{10} (x) = - 6

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

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