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

The answer to ln(x)+x-2=0 is x=\W_{0}(e^2)

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ln(x)+x-2=0

ln (x) + x - 2 = 0

x = W_{0} (e^{2})

Decimal

x = 1.55714 \dots

Solution steps

ln (x) + x - 2 = 0

Move

x

to the right side

ln (x) - 2 = - x

Move

2

to the right side

ln (x) = - x + 2

f (x) = g (x),

then

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

e^{l n (x)} = e^{- x + 2}

Simplify

e^{l n (x)} : x

x = e^{- x + 2}

Solve

x = e^{- x + 2} : x = W_{0} (e^{2})

x = W_{0} (e^{2})

Verify Solutions:

x = W_{0} (e^{2})

True

The solution is

x = W_{0} (e^{2})

ln (x^{3}) = ln (6 x^{2})

ln (2 x + 3) = 5.8

lo g_{2} (12 - 6 x) = 3 lo g_{2} (3)

y, 2.16 = lo g_{10} (x) + y 1.86

lo g_{2} (x + 4) = 5