What is the answer to solvefor x,(x/(10))-a*e^{-x/((r*t))}=0 ?

The answer to solvefor x,(x/(10))-a*e^{-x/((r*t))}=0 is x=\W((10a)/(rt))rt

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solvefor x,(x/(10))-ae^{-x/((rt))}=0

solve for

x, (\frac{x}{10}) - a \cdot e^{- \frac{x}{( r \cdot t )}} = 0

x = W (\frac{10 a}{r t}) r t

Solution steps

(\frac{x}{10}) - a e^{- \frac{x}{( r t )}} = 0

Prepare

(\frac{x}{10}) - a \cdot e^{- \frac{x}{( r \cdot t )}} = 0

for Lambert form:

- 10 a = - e^{\frac{x}{r t}} x

Rewrite the equation with

\frac{x}{r t} = u

and

x = u r t

- 10 a = - e^{u} u r t

Rewrite

- 10 a = - e^{u} u r t

in Lambert form:

e^{u} u = \frac{10 a}{r t}

Solve

e^{u} u = \frac{10 a}{r t} : u = W (\frac{10 a}{r t})

u = W (\frac{10 a}{r t})

Substitute back

u = \frac{x}{r t},

solve for

x

Solve

\frac{x}{r t} = W (\frac{10 a}{r t}) : x = W (\frac{10 a}{r t}) r t

x = W (\frac{10 a}{r t}) r t

n = 1 + (\frac{9}{10})^{n}

(\frac{3}{5})^{1 - 2 x} = \frac{417}{27}

2^{2} x + 3 + 2^{x} + 3 = 1 + 2^{x}

(\frac{1}{32})^{x + 2} = 8^{x + 1}

2^{x} - 1 + 2^{x} + 2^{x} + 1 = 7

What is the answer to solvefor x,(x/(10))-a*e^{-x/((r*t))}=0 ?
The answer to solvefor x,(x/(10))-a*e^{-x/((r*t))}=0 is x=\W((10a)/(rt))rt