What is the answer to 11=sqrt(ln(0^2+x)) ?

The answer to 11=sqrt(ln(0^2+x)) is x=e^{121}

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11=sqrt(ln(0^2+x))

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

x = e^{121}

Decimal Notation

x = 3.54513 E 52

Solution steps

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

Rewrite the equation with

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

11 = u

Solve

11 = u : u = 121

u = 121

Substitute back

u = ln (0^{2} + x),

solve for

x

Solve

ln (0^{2} + x) = 121 : x = e^{121}

x = e^{121}

Verify Solutions:

x = e^{121}

True

The solution is

x = e^{121}

lo g_{10} (x) + lo g_{10} (x + 99) = 2

((ln (y) - 1)^{2}) = 0

3 lo g_{10} (x) = 8 lo g_{10} (x)

lo g_{2.718} (a) = 7.045

lo g_{10} (x) = - lo g_{10} (a) b

What is the answer to 11=sqrt(ln(0^2+x)) ?
The answer to 11=sqrt(ln(0^2+x)) is x=e^{121}
Solve for x: 11=sqrt(ln(0^2+x))
The solution is x=e^{121}