What is the answer to n+20*102^n=24 ?

The answer to n+20*102^n=24 is n=-(\W_{0}(20*102^{24}ln(102))-24ln(102))/(ln(102))

Solve for n: n+20*102^n=24

The solution is n=-(\W_{0}(20*102^{24}ln(102))-24ln(102))/(ln(102))

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n + 20 \cdot 10 2^{n} = 24

n = - \frac{W _{0} ( 20 \cdot 10 2 ^{24} ln ( 102 ) ) - 24 ln ( 102 )}{ln ( 102 )}

Decimal Notation

n = 0.03906 \dots

Solution steps

n + 20 \cdot 10 2^{n} = 24

Prepare

n + 20 \cdot 10 2^{n} = 24

for Lambert form:

20 = (24 - n) e^{- l n (102) n}

Rewrite the equation with

(- n + 24) ln (102) = u

and

n = - \frac{u - 24 ln ( 102 )}{ln ( 102 )}

20 = (24 - (- \frac{u - 24 ln ( 102 )}{ln ( 102 )})) e^{- l n (102) (- \frac{u - 24 l n ( 102 )}{l n ( 102 )})}

Simplify

20 = e^{u - 24 l n (102)} (24 + \frac{u - 24 ln ( 102 )}{ln ( 102 )})

Rewrite

20 = e^{u - 24 l n (102)} (24 + \frac{u - 24 ln ( 102 )}{ln ( 102 )})

in Lambert form:

e^{u} u = 20 \cdot 10 2^{24} ln (102)

Solve

e^{u} u = 20 \cdot 10 2^{24} ln (102) : u = W_{0} (20 \cdot 10 2^{24} ln (102))

u = W_{0} (20 \cdot 10 2^{24} ln (102))

Substitute back

u = (- n + 24) ln (102),

solve for

n

Solve

(- n + 24) ln (102) = W_{0} (20 \cdot 10 2^{24} ln (102)) : n = - \frac{W _{0} ( 20 \cdot 10 2 ^{24} ln ( 102 ) ) - 24 ln ( 102 )}{ln ( 102 )}

n = - \frac{W _{0} ( 20 \cdot 10 2 ^{24} ln ( 102 ) ) - 24 ln ( 102 )}{ln ( 102 )}

\frac{1}{49} = 7^{3 x}

5^{2 x - 8} = 7^{2 x - 8}

(2)^{x} = 42

75 = 1.0 1^{x}

n, (1 + i)^{n} = i

What is the answer to n+20*102^n=24 ?
The answer to n+20*102^n=24 is n=-(\W_{0}(20*102^{24}ln(102))-24ln(102))/(ln(102))
Solve for n: n+20*102^n=24
The solution is n=-(\W_{0}(20*102^{24}ln(102))-24ln(102))/(ln(102))