What is the answer to n=1.2^n ?

The answer to n=1.2^n is n=-(\W_{-1}(-ln(1.2)))/(ln(1.2)),n=-(\W_{0}(-ln(1.2)))/(ln(1.2))

The solution is n=-(\W_{-1}(-ln(1.2)))/(ln(1.2)),n=-(\W_{0}(-ln(1.2)))/(ln(1.2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

n=1.2^n

n = 1. 2^{n}

n = - \frac{W _{- 1} ( - ln ( 1.2 ) )}{ln ( 1.2 )}, n = - \frac{W _{0} ( - ln ( 1.2 ) )}{ln ( 1.2 )}

Solution steps

n = 1. 2^{n}

Prepare

n = 1. 2^{n}

for Lambert form:

n e^{- l n (1.2) n} = 1

Rewrite the equation with

- n ln (1.2) = u

and

n = - \frac{u}{ln ( 1.2 )}

(- \frac{u}{ln ( 1.2 )}) e^{u} = 1

Rewrite

(- \frac{u}{ln ( 1.2 )}) e^{u} = 1

in Lambert form:

e^{u} u = - ln (1.2)

Solve

e^{u} u = - ln (1.2) : u = W_{- 1} (- ln (1.2)), u = W_{0} (- ln (1.2))

u = W_{- 1} (- ln (1.2)), u = W_{0} (- ln (1.2))

Substitute back

u = - n ln (1.2),

solve for

n

Solve

- n ln (1.2) = W_{- 1} (- ln (1.2)) : n = - \frac{W _{- 1} ( - ln ( 1.2 ) )}{ln ( 1.2 )}

Solve

- n ln (1.2) = W_{0} (- ln (1.2)) : n = - \frac{W _{0} ( - ln ( 1.2 ) )}{ln ( 1.2 )}

n = - \frac{W _{- 1} ( - ln ( 1.2 ) )}{ln ( 1.2 )}, n = - \frac{W _{0} ( - ln ( 1.2 ) )}{ln ( 1.2 )}

2^{x} = 3.6 \cdot 1 0^{13}

9^{x - 2} = 240 + 9^{x}

5^{x} - 1 = 3

2.5 \cdot e^{40 x} = 5

(((5^{5})^{x})^{+})^{2} = 25

What is the answer to n=1.2^n ?
The answer to n=1.2^n is n=-(\W_{-1}(-ln(1.2)))/(ln(1.2)),n=-(\W_{0}(-ln(1.2)))/(ln(1.2))
Solve for n: n=1.2^n
The solution is n=-(\W_{-1}(-ln(1.2)))/(ln(1.2)),n=-(\W_{0}(-ln(1.2)))/(ln(1.2))