What is the answer to 22=ln(x)+x ?

The answer to 22=ln(x)+x is x=\W_{0}(e^{22})

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22=ln(x)+x

22 = ln (x) + x

x = W_{0} (e^{22})

Decimal Notation

x = 19.05278 \dots

Solution steps

22 = ln (x) + x

Switch sides

ln (x) + x = 22

Move

x

to the right side

ln (x) = 22 - x

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

e^{l n (x)} = e^{22 - x}

Simplify

e^{l n (x)} : x

x = e^{22 - x}

Solve

x = e^{22 - x} : x = W_{0} (e^{22})

x = W_{0} (e^{22})

Verify Solutions:

x = W_{0} (e^{22})

True

The solution is

x = W_{0} (e^{22})

5 lo g_{10} (x) = 3 \cdot 3

lo g_{x} (3) + 16 lo g_{81} (x) = 5

lo g_{2} (9 x + 2) = 2

lo g_{10} (x) = lo g_{10} (x + 1)

2 ln (x) - ln (3 x - 2) = 0