What is the answer to -x=ln((0.02)/(1+x)) ?

The answer to -x=ln((0.02)/(1+x)) is x=-\W_{-1}(-(0.02)/e)-1,x=-\W_{0}(-(0.02)/e)-1

Solve for x: -x=ln((0.02)/(1+x))

The solution is x=-\W_{-1}(-(0.02)/e)-1,x=-\W_{0}(-(0.02)/e)-1

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- x = ln (\frac{0.02}{1 + x})

x = - W_{- 1} (- \frac{0.02}{e}) - 1, x = - W_{0} (- \frac{0.02}{e}) - 1

Decimal

x = 5.83392 \dots, x = - 0.99258 \dots

Solution steps

- x = ln (\frac{0.02}{1 + x})

f (x) = g (x),

then

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

e^{- x} = e^{l n (\frac{0.02}{1 + x})}

Simplify

e^{l n (\frac{0.02}{1 + x})} : \frac{0.02}{1 + x}

e^{- x} = \frac{0.02}{1 + x}

Multiply both sides by

1 + x

e^{- x} (1 + x) = \frac{0.02}{1 + x} (1 + x)

Simplify

e^{- x} (1 + x) = 0.02

Solve

e^{- x} (1 + x) = 0.02 : x = - W_{- 1} (- \frac{0.02}{e}) - 1, x = - W_{0} (- \frac{0.02}{e}) - 1

x = - W_{- 1} (- \frac{0.02}{e}) - 1, x = - W_{0} (- \frac{0.02}{e}) - 1

Verify Solutions:

x = - W_{- 1} (- \frac{0.02}{e}) - 1

True

, x = - W_{0} (- \frac{0.02}{e}) - 1

True

The solutions are

x = - W_{- 1} (- \frac{0.02}{e}) - 1, x = - W_{0} (- \frac{0.02}{e}) - 1

3 + 2 ln (x) = 10

ln (x + 1) = 2 + ln (x)

x, 5 x + x ln (x) = 0

lo g_{2} (x^{2} - 3) = - 3

ln (3 x) = 2 x - 5

What is the answer to -x=ln((0.02)/(1+x)) ?
The answer to -x=ln((0.02)/(1+x)) is x=-\W_{-1}(-(0.02)/e)-1,x=-\W_{0}(-(0.02)/e)-1
Solve for x: -x=ln((0.02)/(1+x))
The solution is x=-\W_{-1}(-(0.02)/e)-1,x=-\W_{0}(-(0.02)/e)-1