What is the answer to (n+4)/(3^n)= 2/3 ?

The answer to (n+4)/(3^n)= 2/3 is n=2,n=-(\W_{0}(-\frac{2ln(3))/(243))+4ln(3)}{ln(3)}

Solve for n: (n+4)/(3^n)= 2/3

The solution is n=2,n=-(\W_{0}(-\frac{2ln(3))/(243))+4ln(3)}{ln(3)}

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Popular >

(n+4)/(3^n)= 2/3

Solution

\frac{n + 4}{3 ^{n}} = \frac{2}{3}

Solution

n = 2, n = - \frac{W _{0} ( - \frac{2 l n ( 3 )}{243} ) + 4 ln ( 3 )}{ln ( 3 )}

Decimal Notation

n = 2, n = - 3.99169 \dots

Solution steps

\frac{n + 4}{3 ^{n}} = \frac{2}{3}

Prepare

\frac{n + 4}{3 ^{n}} = \frac{2}{3}

for Lambert form:

(n + 4) e^{- l n (3) n} = \frac{2}{3}

Rewrite the equation with

(- n - 4) ln (3) = u

and

n = - \frac{u + 4 ln ( 3 )}{ln ( 3 )}

((- \frac{u + 4 ln ( 3 )}{ln ( 3 )}) + 4) e^{- l n (3) (- \frac{u + 4 l n ( 3 )}{l n ( 3 )})} = \frac{2}{3}

Simplify

e^{u + 4 l n (3)} (- \frac{u + 4 ln ( 3 )}{ln ( 3 )} + 4) = \frac{2}{3}

Rewrite

e^{u + 4 l n (3)} (- \frac{u + 4 ln ( 3 )}{ln ( 3 )} + 4) = \frac{2}{3}

in Lambert form:

e^{u} u = - \frac{2 ln ( 3 )}{243}

Solve

e^{u} u = - \frac{2 ln ( 3 )}{243} : u = - 6 ln (3), u = W_{0} (- \frac{2 ln ( 3 )}{243})

u = - 6 ln (3), u = W_{0} (- \frac{2 ln ( 3 )}{243})

Substitute back

u = (- n - 4) ln (3),

solve for

n

Solve

(- n - 4) ln (3) = - 6 ln (3) : n = 2

Solve

(- n - 4) ln (3) = W_{0} (- \frac{2 ln ( 3 )}{243}) : n = - \frac{W _{0} ( - \frac{2 l n ( 3 )}{243} ) + 4 ln ( 3 )}{ln ( 3 )}

n = 2, n = - \frac{W _{0} ( - \frac{2 l n ( 3 )}{243} ) + 4 ln ( 3 )}{ln ( 3 )}

Verify Solutions:

n = 2

True

, n = - \frac{W _{0} ( - \frac{2 l n ( 3 )}{243} ) + 4 ln ( 3 )}{ln ( 3 )}

True

The solutions are

n = 2, n = - \frac{W _{0} ( - \frac{2 l n ( 3 )}{243} ) + 4 ln ( 3 )}{ln ( 3 )}

Verify Solutions:

n = 2

True

, n = - \frac{W _{0} ( - \frac{2 l n ( 3 )}{243} ) + 4 ln ( 3 )}{ln ( 3 )}

True

The solutions are

n = 2, n = - \frac{W _{0} ( - \frac{2 l n ( 3 )}{243} ) + 4 ln ( 3 )}{ln ( 3 )}

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

2=16^{0.0125t}

2 = 1 6^{0.0125 t}

16^{x+4}=64^{3x-5}

1 6^{x + 4} = 6 4^{3 x - 5}

Frequently Asked Questions (FAQ)

What is the answer to (n+4)/(3^n)= 2/3 ?
The answer to (n+4)/(3^n)= 2/3 is n=2,n=-(\W_{0}(-\frac{2ln(3))/(243))+4ln(3)}{ln(3)}
Solve for n: (n+4)/(3^n)= 2/3
The solution is n=2,n=-(\W_{0}(-\frac{2ln(3))/(243))+4ln(3)}{ln(3)}