What is ln((a^1)/(b^1c^{-1)}) ?

The solution to ln((a^1)/(b^1c^{-1)}) is ln(a)-ln(b)+ln(c)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify ln((a^1)/(b^1c^{-1)})

simplify

ln (\frac{a ^{1}}{b ^{1} c ^{- 1}})

ln (a) - ln (b) + ln (c)

Solution steps

ln (\frac{a ^{1}}{b ^{1} c ^{- 1}})

Apply log rule:

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

ln (\frac{a ^{1}}{b ^{1} c ^{- 1}}) = ln (a^{1}) - ln (b^{1} c^{- 1})

= ln (a^{1}) - ln (b^{1} c^{- 1})

Apply log rule:

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

ln (a^{1}) = 1 \cdot ln (a)

= 1 \cdot ln (a) - ln (b^{1} c^{- 1})

Apply log rule:

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

ln (b^{1} c^{- 1}) = ln (b^{1}) + ln (c^{- 1})

= 1 \cdot ln (a) - (ln (b^{1}) + ln (c^{- 1}))

Apply log rule:

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

ln (b^{1}) = 1 \cdot ln (b)

= 1 \cdot ln (a) - (1 \cdot ln (b) + ln (c^{- 1}))

Apply log rule:

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

ln (c^{- 1}) = - 1 \cdot ln (c)

= 1 \cdot ln (a) - (1 \cdot ln (b) - 1 \cdot ln (c))

1 \cdot ln (a) = ln (a)

- (1 \cdot ln (b) - 1 \cdot ln (c)) = - (ln (b) - ln (c))

= ln (a) - (ln (b) - ln (c))

- (ln (b) - ln (c)) : - ln (b) + ln (c)

= ln (a) - ln (b) + ln (c)

- lo g_{10} (3.6 \cdot 1 0^{- 10})

2 ln (x) - 3 ln (z) - ln (w)

\frac{ln ( 5 )}{ln ( 5 ) - 2 ln ( 2 )}

ln (3 x) - ln (5 y)

38 ln (\frac{90}{10})

What is ln((a^1)/(b^1c^{-1)}) ?
The solution to ln((a^1)/(b^1c^{-1)}) is ln(a)-ln(b)+ln(c)