What is ln((v^2)/(2αe^{-βs)}) ?

The solution to ln((v^2)/(2αe^{-βs)}) is 2ln(v)-ln(2)-ln(α)+βs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify ln((v^2)/(2αe^{-βs)})

simplify

ln (\frac{v ^{2}}{2 α e ^{- β s}})

2 ln (v) - ln (2) - ln (α) + β s

Solution steps

ln (\frac{v ^{2}}{2 α e ^{- β s}})

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

ln (\frac{v ^{2}}{2 α e ^{- β s}}) = ln (v^{2}) - ln (2 α e^{- β s})

= ln (v^{2}) - ln (2 α e^{- β s})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (v^{2}) = 2 ln (v)

= 2 ln (v) - ln (2 α e^{- β s})

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

ln (2 α e^{- β s}) = ln (2) + ln (α) + ln (e^{- β s})

= 2 ln (v) - (ln (e^{- β s}) + ln (α) + ln (2))

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (e^{- β s}) = - β s ln (e)

= 2 ln (v) - (ln (2) + ln (α) - β s ln (e))

β s ln (e) = β s

= 2 ln (v) - (- β s + ln (α) + ln (2))

- (ln (2) + ln (α) - β s) : - ln (2) - ln (α) + β s

= 2 ln (v) - ln (2) - ln (α) + β s

lo g_{10} (\frac{3 x}{y ^{3}})

\frac{1}{- 0.20698} ln (\frac{28}{118})

\frac{1}{2} ln (\frac{x ^{2} + ( y - 1 ) ^{2}}{x ^{2}}) + ln (x)

2 ln (+ x^{3} ln (x))

ln (3 + 2 y) - ln (1 + 3 y)

What is ln((v^2)/(2αe^{-βs)}) ?
The solution to ln((v^2)/(2αe^{-βs)}) is 2ln(v)-ln(2)-ln(α)+βs