What is 1/2 (-1/3 ln|y|+1/3 ln|y-3|) ?

The solution to 1/2 (-1/3 ln|y|+1/3 ln|y-3|) is 1/2 ln(\sqrt[3]{(|y-3|)/(|y|)})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify 1/2 (-1/3 ln|y|+1/3 ln|y-3|)

simplify

\frac{1}{2} (- \frac{1}{3} ln ∣ y ∣ + \frac{1}{3} ln ∣ y - 3 ∣)

Solution steps

\frac{1}{2} (- \frac{1}{3} ln ∣ y ∣ + \frac{1}{3} ln ∣ y - 3 ∣)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

\frac{1}{3} ln ∣ y ∣ = ln (∣ y ∣^{\frac{1}{3}})

= \frac{1}{2} (\frac{1}{3} ln ∣ y - 3 ∣ - ln (∣ y ∣^{\frac{1}{3}}))

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

\frac{1}{3} ln ∣ y - 3 ∣ = ln (∣ y - 3 ∣^{\frac{1}{3}})

= \frac{1}{2} (ln (∣ y - 3 ∣^{\frac{1}{3}}) - ln (∣ y ∣^{\frac{1}{3}}))

Apply log rule:

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

ln (∣ y - 3 ∣^{\frac{1}{3}}) - ln (∣ y ∣^{\frac{1}{3}}) = ln (\frac{∣ y - 3 ∣ ^{\frac{1}{3}}}{∣ y ∣ ^{\frac{1}{3}}})

= \frac{1}{2} (ln (\frac{∣ y - 3 ∣ ^{\frac{1}{3}}}{∣ y ∣ ^{\frac{1}{3}}}))

Combine same powers :

Remove parentheses:

(a) = a

lo g_{10} (10 x) + lo g_{10} (20 y) - lo g_{10} (10)

ln (\frac{23}{20})

ln (c \frac{x ^{a}}{z ^{b}})

\frac{1}{2} ln (\frac{120}{500})

ln (5) + \frac{1}{3} ln (8)

What is 1/2 (-1/3 ln|y|+1/3 ln|y-3|) ?
The solution to 1/2 (-1/3 ln|y|+1/3 ln|y-3|) is 1/2 ln(\sqrt[3]{(|y-3|)/(|y|)})