What is 1/4 ln|x|-1/8 ln|x^2+4| ?

The solution to 1/4 ln|x|-1/8 ln|x^2+4| is ln((|x|^{1/4}|x^2+4|^{7/8})/(|x^2+4|))

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simplify 1/4 ln|x|-1/8 ln|x^2+4|

simplify

\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4

ln \frac{∣ x ∣ ^{\frac{1}{4}} x ^{2} + 4 ^{\frac{7}{8}}}{∣ x ^{2} + 4 ∣}

Solution steps

\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4

Apply log rule:

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

\frac{1}{4} ln ∣ x ∣ = ln (∣ x ∣^{\frac{1}{4}})

= ln (∣ x ∣^{\frac{1}{4}}) - \frac{1}{8} ln x^{2} + 4

Apply log rule:

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

\frac{1}{8} ln x^{2} + 4 = ln (x^{2} + 4^{\frac{1}{8}})

= ln (∣ x ∣^{\frac{1}{4}}) - ln (x^{2} + 4^{\frac{1}{8}})

Apply log rule:

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

ln (∣ x ∣^{\frac{1}{4}}) - ln (x^{2} + 4^{\frac{1}{8}}) = ln (\frac{∣ x ∣ ^{\frac{1}{4}}}{∣ x ^{2} + 4 ∣ ^{\frac{1}{8}}})

= ln (\frac{∣ x ∣ ^{\frac{1}{4}}}{∣ x ^{2} + 4 ∣ ^{\frac{1}{8}}})

Simplify

\frac{∣ x ∣ ^{\frac{1}{4}}}{∣ x ^{2} + 4 ∣ ^{\frac{1}{8}}} : \frac{∣ x ∣ ^{\frac{1}{4}} x ^{2} + 4 ^{\frac{7}{8}}}{∣ x ^{2} + 4 ∣}

= ln \frac{∣ x ∣ ^{\frac{1}{4}} x ^{2} + 4 ^{\frac{7}{8}}}{∣ x ^{2} + 4 ∣}

\frac{1}{2} ln (e^{m x} e^{m y} (1 - z^{2})^{\frac{m}{2}})

lo g_{10} (\frac{x}{y ^{8}})

- ln (y) + ln (y + 1)

- 12 ln (8) + 12 ln (4)

- \frac{3}{4} lo g_{2} (\frac{3}{4}) - \frac{1}{4} lo g_{10} (x)

What is 1/4 ln|x|-1/8 ln|x^2+4| ?
The solution to 1/4 ln|x|-1/8 ln|x^2+4| is ln((|x|^{1/4}|x^2+4|^{7/8})/(|x^2+4|))