What is the answer to 1.02^n+50*n=-200 ?

The answer to 1.02^n+50*n=-200 is n=-(\W_{0}(\frac{ln(1.02))/(54.121608))+4ln(1.02)}{ln(1.02)}

Solve for n: 1.02^n+50*n=-200

The solution is n=-(\W_{0}(\frac{ln(1.02))/(54.121608))+4ln(1.02)}{ln(1.02)}

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1.0 2^{n} + 50 \cdot n = - 200

n = - \frac{W _{0} ( \frac{l n ( 1.02 )}{54.121608} ) + 4 ln ( 1.02 )}{ln ( 1.02 )}

Decimal Notation

n = - 4.01847 \dots

Solution steps

1.0 2^{n} + 50 n = - 200

Prepare

1.0 2^{n} + 50 \cdot n = - 200

for Lambert form:

1 = (- 200 - 50 n) e^{- l n (1.02) n}

Rewrite the equation with

(- n - 4) ln (1.02) = u

and

n = - \frac{u + 4 ln ( 1.02 )}{ln ( 1.02 )}

1 = (- 200 - 50 (- \frac{u + 4 ln ( 1.02 )}{ln ( 1.02 )})) e^{- l n (1.02) (- \frac{u + 4 l n ( 1.02 )}{l n ( 1.02 )})}

Simplify

1 = e^{u + 4 l n (1.02)} (- 200 + \frac{50 ( u + 4 ln ( 1.02 ) )}{ln ( 1.02 )})

Rewrite

1 = e^{u + 4 l n (1.02)} (- 200 + \frac{50 ( u + 4 ln ( 1.02 ) )}{ln ( 1.02 )})

in Lambert form:

e^{u} u = \frac{ln ( 1.02 )}{54.121608}

Solve

e^{u} u = \frac{ln ( 1.02 )}{54.121608} : u = W_{0} (\frac{ln ( 1.02 )}{54.121608})

u = W_{0} (\frac{ln ( 1.02 )}{54.121608})

Substitute back

u = (- n - 4) ln (1.02),

solve for

n

Solve

(- n - 4) ln (1.02) = W_{0} (\frac{ln ( 1.02 )}{54.121608}) : n = - \frac{W _{0} ( \frac{l n ( 1.02 )}{54.121608} ) + 4 ln ( 1.02 )}{ln ( 1.02 )}

n = - \frac{W _{0} ( \frac{l n ( 1.02 )}{54.121608} ) + 4 ln ( 1.02 )}{ln ( 1.02 )}

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2^{3 x - 13} = \frac{1}{16}

(e^{x})^{2} = e^{x}

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2 5^{3 x + 1} = 12 5^{2 x - 1}

What is the answer to 1.02^n+50*n=-200 ?
The answer to 1.02^n+50*n=-200 is n=-(\W_{0}(\frac{ln(1.02))/(54.121608))+4ln(1.02)}{ln(1.02)}
Solve for n: 1.02^n+50*n=-200
The solution is n=-(\W_{0}(\frac{ln(1.02))/(54.121608))+4ln(1.02)}{ln(1.02)}