What is simplify-1/3 ln|1-3v^2|-ln|x| ?

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

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simplify-1/3 ln|1-3v^2|-ln|x|

simplify

- \frac{1}{3} ln 1 - 3 v^{2} - ln ∣ x ∣

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

Solution steps

- \frac{1}{3} ln 1 - 3 v^{2} - ln ∣ x ∣

Apply log rule:

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

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

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

Rewrite

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

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

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

Apply log rule:

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

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

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

- lo g_{10} (5.159 \cdot 1 0^{13})

ln (\frac{1}{b ^{2 n}} \cdot e^{- \frac{x}{nb}} \cdot x)

- \frac{1}{2} ln (\frac{0.0009}{2})

ln (1 + \frac{1}{x}) - ln (1 + \frac{1}{x - 1})

- \frac{24 ln ( 2 ) - 3 ln ( 15 )}{2 ( ln ( 15 ) - 9 ln ( 2 ) )}

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