What is the answer to ln(ln(x^2+1))=0 ?

The answer to ln(ln(x^2+1))=0 is x=sqrt(e-1),x=-sqrt(e-1)

Solve for x: ln(ln(x^2+1))=0

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

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ln (ln (x^{2} + 1)) = 0

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

Decimal Notation

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

Solution steps

ln (ln (x^{2} + 1)) = 0

Apply log rules

x^{2} + 1 = e

Solve

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

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

Verify Solutions:

x = e - 1

True

, x = - e - 1

True

The solutions are

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

x, ln (y) = ln (x) - x + c

ln (x) - 5 = - 3

ln (x) + ln (x + 1) = ln (6)

x, ln (y) = 5 ln (18 + x)

4 lo g_{6} (2) = lo g_{6} (x)

What is the answer to ln(ln(x^2+1))=0 ?
The answer to ln(ln(x^2+1))=0 is x=sqrt(e-1),x=-sqrt(e-1)
Solve for x: ln(ln(x^2+1))=0
The solution is x=sqrt(e-1),x=-sqrt(e-1)