What is the answer to 250=((e^{42r}-1))/r ?

The answer to 250=((e^{42r}-1))/r is r=-(125\W_{-1}(-\frac{21)/(125e^{21/125)})+21}{5250}

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250=((e^{42r}-1))/r

250 = \frac{( e ^{42 r} - 1 )}{r}

r = - \frac{125 W _{- 1} ( - \frac{21}{125 e ^{\frac{21}{125}}} ) + 21}{5250}

Decimal Notation

r = 0.06921 \dots

Solution steps

250 = \frac{( e ^{42 r} - 1 )}{r}

Prepare

250 = \frac{( e ^{42 r} - 1 )}{r}

for Lambert form:

1 = (250 r + 1) e^{- 42 r}

Rewrite the equation with

- 42 r - \frac{21}{125} = u

and

r = - \frac{125 u + 21}{5250}

1 = (250 (- \frac{125 u + 21}{5250}) + 1) e^{- 42 (- \frac{125 u + 21}{5250})}

Simplify

1 = e^{\frac{125 u + 21}{125}} (- \frac{125 u + 21}{21} + 1)

Rewrite

1 = e^{\frac{125 u + 21}{125}} (- \frac{125 u + 21}{21} + 1)

in Lambert form:

e^{u} u = - \frac{21}{125 e ^{\frac{21}{125}}}

Solve

e^{u} u = - \frac{21}{125 e ^{\frac{21}{125}}} : u = W_{- 1} (- \frac{21}{125 e ^{\frac{21}{125}}}), u = W_{0} (- \frac{21}{125 e ^{\frac{21}{125}}})

u = W_{- 1} (- \frac{21}{125 e ^{\frac{21}{125}}}), u = W_{0} (- \frac{21}{125 e ^{\frac{21}{125}}})

Substitute back

u = - 42 r - \frac{21}{125},

solve for

r

Solve

- 42 r - \frac{21}{125} = W_{- 1} (- \frac{21}{125 e ^{\frac{21}{125}}}) : r = - \frac{125 W _{- 1} ( - \frac{21}{125 e ^{\frac{21}{125}}} ) + 21}{5250}

Solve

- 42 r - \frac{21}{125} = W_{0} (- \frac{21}{125 e ^{\frac{21}{125}}}) : r = - \frac{125 W _{0} ( - \frac{21}{125 e ^{\frac{21}{125}}} ) + 21}{5250}

r = - \frac{125 W _{- 1} ( - \frac{21}{125 e ^{\frac{21}{125}}} ) + 21}{5250}, r = - \frac{125 W _{0} ( - \frac{21}{125 e ^{\frac{21}{125}}} ) + 21}{5250}

Verify Solutions:

r = - \frac{125 W _{- 1} ( - \frac{21}{125 e ^{\frac{21}{125}}} ) + 21}{5250}

True

, r = - \frac{125 W _{0} ( - \frac{21}{125 e ^{\frac{21}{125}}} ) + 21}{5250}

False

The solution is

r = - \frac{125 W _{- 1} ( - \frac{21}{125 e ^{\frac{21}{125}}} ) + 21}{5250}

Verify Solutions:

r = - \frac{125 W _{- 1} ( - \frac{21}{125 e ^{\frac{21}{125}}} ) + 21}{5250}

True

The solution is

r = - \frac{125 W _{- 1} ( - \frac{21}{125 e ^{\frac{21}{125}}} ) + 21}{5250}

(5^{2})^{x} = 125

9^{x^{2}} \cdot 3^{7 x} = 81

5^{x} = 2^{x + 3}

4^{x} + 2^{x} - 2 = 0

21 6^{x} = 26

What is the answer to 250=((e^{42r}-1))/r ?
The answer to 250=((e^{42r}-1))/r is r=-(125\W_{-1}(-\frac{21)/(125e^{21/125)})+21}{5250}