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

The answer to 0=ln(x)-1/2 x+1 is x=-2\W_{-1}(-1/(2e)),x=-2\W_{0}(-1/(2e))

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

The solution is x=-2\W_{-1}(-1/(2e)),x=-2\W_{0}(-1/(2e))

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0 = ln (x) - \frac{1}{2} x + 1

x = - 2 W_{- 1} (- \frac{1}{2 e}), x = - 2 W_{0} (- \frac{1}{2 e})

Solution steps

0 = ln (x) - \frac{1}{2} x + 1

Switch sides

ln (x) - \frac{1}{2} x + 1 = 0

Move

\frac{1}{2} x

to the right side

ln (x) + 1 = \frac{1}{2} x

Move

1

to the right side

ln (x) = \frac{1}{2} x - 1

f (x) = g (x),

then

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

e^{l n (x)} = e^{\frac{1}{2} x - 1}

Simplify

e^{l n (x)} : x

x = e^{\frac{1}{2} x - 1}

Solve

x = e^{\frac{1}{2} x - 1} : x = - 2 W_{- 1} (- \frac{1}{2 e}), x = - 2 W_{0} (- \frac{1}{2 e})

x = - 2 W_{- 1} (- \frac{1}{2 e}), x = - 2 W_{0} (- \frac{1}{2 e})

Verify Solutions:

x = - 2 W_{- 1} (- \frac{1}{2 e})

True

, x = - 2 W_{0} (- \frac{1}{2 e})

True

The solutions are

x = - 2 W_{- 1} (- \frac{1}{2 e}), x = - 2 W_{0} (- \frac{1}{2 e})

3 ln (x) - 1 = 0

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

lo g_{10} (x) 343 = 3

ln (x) = - 9

lo g_{10} (3 x) + lo g_{10} (x + 2) = 0

What is the answer to 0=ln(x)-1/2 x+1 ?
The answer to 0=ln(x)-1/2 x+1 is x=-2\W_{-1}(-1/(2e)),x=-2\W_{0}(-1/(2e))
Solve for x: 0=ln(x)-1/2 x+1
The solution is x=-2\W_{-1}(-1/(2e)),x=-2\W_{0}(-1/(2e))