What is the answer to d(x)=5(1/3)^x ?

The answer to d(x)=5(1/3)^x is x=(\W((5ln(3))/d))/(ln(3))

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d(x)=5(1/3)^x

d (x) = 5 (\frac{1}{3})^{x}

x = \frac{W ( \frac{5 l n ( 3 )}{d} )}{ln ( 3 )}

Solution steps

d (x) = 5 (\frac{1}{3})^{x}

Prepare

d (x) = 5 (\frac{1}{3})^{x}

for Lambert form:

d x e^{l n (3) x} = 5

Rewrite the equation with

ln (3) x = u

and

x = \frac{u}{ln ( 3 )}

d (\frac{u}{ln ( 3 )}) e^{u} = 5

Rewrite

d (\frac{u}{ln ( 3 )}) e^{u} = 5

in Lambert form:

e^{u} u = \frac{5 ln ( 3 )}{d}

Solve

e^{u} u = \frac{5 ln ( 3 )}{d} : u = W (\frac{5 ln ( 3 )}{d})

u = W (\frac{5 ln ( 3 )}{d})

Substitute back

u = ln (3) x,

solve for

x

Solve

ln (3) x = W (\frac{5 ln ( 3 )}{d}) : x = \frac{W ( \frac{5 l n ( 3 )}{d} )}{ln ( 3 )}

x = \frac{W ( \frac{5 l n ( 3 )}{d} )}{ln ( 3 )}

y^{^{'''}} + y^{^{''}} = x + e^{- x}

y^{^{''}} - 2 y^{^{'}} + y = e^{2 x} + x + 1

y^{^{''}} - 7 y^{^{'}} + 6 y = (20 x + 11) e^{x}

y^{^{''}} - y^{^{'}} + \frac{1}{4} y = 7 + e^{\frac{x}{2}}

V^{^{''}} + 25 \cdot 1 0^{2} V^{^{'}} + 25 \cdot 1 0^{6} V = 5 \cdot 1 0^{8}

What is the answer to d(x)=5(1/3)^x ?
The answer to d(x)=5(1/3)^x is x=(\W((5ln(3))/d))/(ln(3))