What is simplify-1/2 ln|e^y|+1/2 ln|e^y-2| ?

The solution to simplify-1/2 ln|e^y|+1/2 ln|e^y-2| is ln(sqrt((|e^y-2|)/(e^y)))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify-1/2 ln|e^y|+1/2 ln|e^y-2|

simplify

- \frac{1}{2} ln ∣ e^{y} ∣ + \frac{1}{2} ln ∣ e^{y} - 2 ∣

ln (\frac{∣ e ^{y} - 2 ∣}{e ^{y}})

Solution steps

- \frac{1}{2} ln ∣ e^{y} ∣ + \frac{1}{2} ln ∣ e^{y} - 2 ∣

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Simplify

\frac{∣ e ^{y} - 2 ∣ ^{\frac{1}{2}}}{∣ e ^{y} ∣ ^{\frac{1}{2}}} : \frac{∣ e ^{y} - 2 ∣}{e ^{y}}

= ln (\frac{∣ e ^{y} - 2 ∣}{e ^{y}})

lo g_{10} (w^{6} u \cdot v)

lo g_{3} ((\frac{4 x}{y})^{2})

ln (x^{14} 6 - x)

lo g_{10} (256 x^{8})

ln (x^{2} e^{- 2 x})

What is simplify-1/2 ln|e^y|+1/2 ln|e^y-2| ?
The solution to simplify-1/2 ln|e^y|+1/2 ln|e^y-2| is ln(sqrt((|e^y-2|)/(e^y)))