What is the answer to 6k+ln(1-4k)=0 ?

The answer to 6k+ln(1-4k)=0 is k=(2\W_{-1}(-3/(2e^{3/2)})+3)/(12),k=(2\W_{0}(-3/(2e^{3/2)})+3)/(12)

Solve for k: 6k+ln(1-4k)=0

The solution is k=(2\W_{-1}(-3/(2e^{3/2)})+3)/(12),k=(2\W_{0}(-3/(2e^{3/2)})+3)/(12)

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6 k + ln (1 - 4 k) = 0

k = \frac{2 W _{- 1} ( - \frac{3}{2 e ^{\frac{3}{2}}} ) + 3}{12}, k = \frac{2 W _{0} ( - \frac{3}{2 e ^{\frac{3}{2}}} ) + 3}{12}

Solution steps

6 k + ln (1 - 4 k) = 0

Move

6 k

to the right side

ln (1 - 4 k) = - 6 k

f (x) = g (x),

then

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

e^{l n (1 - 4 k)} = e^{- 6 k}

Simplify

e^{l n (1 - 4 k)} : 1 - 4 k

1 - 4 k = e^{- 6 k}

Solve

1 - 4 k = e^{- 6 k} : k = \frac{2 W _{- 1} ( - \frac{3}{2 e ^{\frac{3}{2}}} ) + 3}{12}, k = \frac{2 W _{0} ( - \frac{3}{2 e ^{\frac{3}{2}}} ) + 3}{12}

k = \frac{2 W _{- 1} ( - \frac{3}{2 e ^{\frac{3}{2}}} ) + 3}{12}, k = \frac{2 W _{0} ( - \frac{3}{2 e ^{\frac{3}{2}}} ) + 3}{12}

Verify Solutions:

k = \frac{2 W _{- 1} ( - \frac{3}{2 e ^{\frac{3}{2}}} ) + 3}{12}

True

, k = \frac{2 W _{0} ( - \frac{3}{2 e ^{\frac{3}{2}}} ) + 3}{12}

True

The solutions are

k = \frac{2 W _{- 1} ( - \frac{3}{2 e ^{\frac{3}{2}}} ) + 3}{12}, k = \frac{2 W _{0} ( - \frac{3}{2 e ^{\frac{3}{2}}} ) + 3}{12}

ln (x) = ln (5) + ln (8)

lo g_{5} (25 x) + lo g_{x} (5) = 4

lo g_{5} (11 - 6 x) = lo g_{5} (1 - x)

2 ln (x - 1) + 6 = 8

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

What is the answer to 6k+ln(1-4k)=0 ?
The answer to 6k+ln(1-4k)=0 is k=(2\W_{-1}(-3/(2e^{3/2)})+3)/(12),k=(2\W_{0}(-3/(2e^{3/2)})+3)/(12)
Solve for k: 6k+ln(1-4k)=0
The solution is k=(2\W_{-1}(-3/(2e^{3/2)})+3)/(12),k=(2\W_{0}(-3/(2e^{3/2)})+3)/(12)