What is ln|x|+1/2 ln|x+2|-1/3 ln|x-3| ?

The solution to ln|x|+1/2 ln|x+2|-1/3 ln|x-3| is ln((|x||x+2|^{1/2}|x-3|^{2/3})/(|x-3|))

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simplify ln|x|+1/2 ln|x+2|-1/3 ln|x-3|

simplify

ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 2 ∣ - \frac{1}{3} ln ∣ x - 3 ∣

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

Solution steps

ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 2 ∣ - \frac{1}{3} ln ∣ x - 3 ∣

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Simplify

\frac{∣ x ∣ ∣ x + 2 ∣ ^{\frac{1}{2}}}{∣ x - 3 ∣ ^{\frac{1}{3}}} : \frac{∣ x ∣ ∣ x + 2 ∣ ^{\frac{1}{2}} ∣ x - 3 ∣ ^{\frac{2}{3}}}{∣ x - 3 ∣}

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

ln (\frac{m}{8} (p - 9))

\frac{1}{3} (6 ln (x + 7) + ln (x) - ln (x^{2} - 8))

- lo g_{10} (1.35 (1 0^{- 10}))

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

ln (\frac{0.27}{2.21 \cdot 1 0 ^{4}}) \cdot \frac{8.314}{90315}

What is ln|x|+1/2 ln|x+2|-1/3 ln|x-3| ?
The solution to ln|x|+1/2 ln|x+2|-1/3 ln|x-3| is ln((|x||x+2|^{1/2}|x-3|^{2/3})/(|x-3|))