What is the answer to c+5ln(-1+c)=0.540302 ?

The answer to c+5ln(-1+c)=0.540302 is c=5\W_{0}(1/(5e^{\frac{0.459698){5}}})+1

Solve for c: c+5ln(-1+c)=0.540302

The solution is c=5\W_{0}(1/(5e^{\frac{0.459698){5}}})+1

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c + 5 ln (- 1 + c) = 0.540302

c = 5 W_{0} (\frac{1}{5 e ^{\frac{0.459698}{5}}}) + 1

Decimal

c = 1.78035 \dots

Solution steps

c + 5 ln (- 1 + c) = 0.540302

Move

c

to the right side

5 ln (- 1 + c) = 0.540302 - c

Divide both sides by

5

ln (- 1 + c) = \frac{0.540302 - c}{5}

f (x) = g (x),

then

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

e^{l n (- 1 + c)} = e^{\frac{0.540302 - c}{5}}

Simplify

e^{l n (- 1 + c)} : - 1 + c

- 1 + c = e^{\frac{0.540302 - c}{5}}

Solve

- 1 + c = e^{\frac{0.540302 - c}{5}} : c = 5 W_{0} (\frac{1}{5 e ^{\frac{0.459698}{5}}}) + 1

c = 5 W_{0} (\frac{1}{5 e ^{\frac{0.459698}{5}}}) + 1

Verify Solutions:

c = 5 W_{0} (\frac{1}{5 e ^{\frac{0.459698}{5}}}) + 1

True

The solution is

c = 5 W_{0} (\frac{1}{5 e ^{\frac{0.459698}{5}}}) + 1

lo g_{8} (x) = \frac{1}{6}

lo g_{3} (x - 9) = 4

lo g_{2} (x) + lo g_{2} (2 x) = lo g_{3} (9)

ln (x) + ln (x + 8) = 3

lo g_{10} (N + M)

What is the answer to c+5ln(-1+c)=0.540302 ?
The answer to c+5ln(-1+c)=0.540302 is c=5\W_{0}(1/(5e^{\frac{0.459698){5}}})+1
Solve for c: c+5ln(-1+c)=0.540302
The solution is c=5\W_{0}(1/(5e^{\frac{0.459698){5}}})+1