What is the answer to n=1+(9/10)^n ?

The answer to n=1+(9/10)^n is n=-(\W_{0}(-\frac{9ln(9/10))/(10))-ln(9/10)}{ln(9/10)}

Solve for n: n=1+(9/10)^n

The solution is n=-(\W_{0}(-\frac{9ln(9/10))/(10))-ln(9/10)}{ln(9/10)}

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

n = - \frac{W _{0} ( - \frac{9 l n ( \frac{9}{10} )}{10} ) - ln ( \frac{9}{10} )}{ln ( \frac{9}{10} )}

Decimal Notation

n = 1.82506 \dots

Solution steps

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

Prepare

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

for Lambert form:

1 = (n - 1) e^{- l n (\frac{9}{10}) n}

Rewrite the equation with

ln (\frac{9}{10}) (- n + 1) = u

and

n = - \frac{u - ln ( \frac{9}{10} )}{ln ( \frac{9}{10} )}

1 = ((- \frac{u - ln ( \frac{9}{10} )}{ln ( \frac{9}{10} )}) - 1) e^{- l n (\frac{9}{10}) (- \frac{u - l n ( \frac{9}{10} )}{l n ( \frac{9}{10} )})}

Simplify

1 = e^{u - l n (\frac{9}{10})} (- \frac{u - ln ( \frac{9}{10} )}{ln ( \frac{9}{10} )} - 1)

Rewrite

1 = e^{u - l n (\frac{9}{10})} (- \frac{u - ln ( \frac{9}{10} )}{ln ( \frac{9}{10} )} - 1)

in Lambert form:

e^{u} u = - \frac{9 ln ( \frac{9}{10} )}{10}

Solve

e^{u} u = - \frac{9 ln ( \frac{9}{10} )}{10} : u = W_{0} (- \frac{9 ln ( \frac{9}{10} )}{10})

u = W_{0} (- \frac{9 ln ( \frac{9}{10} )}{10})

Substitute back

u = ln (\frac{9}{10}) (- n + 1),

solve for

n

Solve

ln (\frac{9}{10}) (- n + 1) = W_{0} (- \frac{9 ln ( \frac{9}{10} )}{10}) : n = - \frac{W _{0} ( - \frac{9 l n ( \frac{9}{10} )}{10} ) - ln ( \frac{9}{10} )}{ln ( \frac{9}{10} )}

n = - \frac{W _{0} ( - \frac{9 l n ( \frac{9}{10} )}{10} ) - ln ( \frac{9}{10} )}{ln ( \frac{9}{10} )}

(\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

3^{2 x - 5} + 9^{x - 2} = 4

What is the answer to n=1+(9/10)^n ?
The answer to n=1+(9/10)^n is n=-(\W_{0}(-\frac{9ln(9/10))/(10))-ln(9/10)}{ln(9/10)}
Solve for n: n=1+(9/10)^n
The solution is n=-(\W_{0}(-\frac{9ln(9/10))/(10))-ln(9/10)}{ln(9/10)}