What is 1/4 ln(0-1)-1/4 ln(0+3) ?

The solution to 1/4 ln(0-1)-1/4 ln(0+3) is 1/4 ln(-1/3)

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simplify 1/4 ln(0-1)-1/4 ln(0+3)

simplify

\frac{1}{4} ln (0 - 1) - \frac{1}{4} ln (0 + 3)

\frac{1}{4} ln (- \frac{1}{3})

Solution steps

\frac{1}{4} ln (0 - 1) - \frac{1}{4} ln (0 + 3)

Apply log rule:

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

\frac{1}{4} ln (0 - 1) = ln ((0 - 1)^{\frac{1}{4}})

= ln ((0 - 1)^{\frac{1}{4}}) - \frac{1}{4} ln (0 + 3)

Apply log rule:

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

\frac{1}{4} ln (0 + 3) = ln ((0 + 3)^{\frac{1}{4}})

= ln ((0 - 1)^{\frac{1}{4}}) - ln ((0 + 3)^{\frac{1}{4}})

Apply log rule:

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

ln ((0 - 1)^{\frac{1}{4}}) - ln ((0 + 3)^{\frac{1}{4}}) = ln (\frac{( 0 - 1 ) ^{\frac{1}{4}}}{( 0 + 3 ) ^{\frac{1}{4}}})

= ln (\frac{( 0 - 1 ) ^{\frac{1}{4}}}{( 0 + 3 ) ^{\frac{1}{4}}})

Simplify

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

Rewrite as

= ln (((- 1) \frac{1}{3})^{\frac{1}{4}})

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

= \frac{1}{4} ln ((- 1) \frac{1}{3})

(- 1) \frac{1}{3} = - \frac{1}{3}

= \frac{1}{4} ln (- \frac{1}{3})

ln (z^{4} e^{2 x})

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

lo g_{10} (\frac{5 \cdot 1 0 ^{- 6}}{1 \cdot 1 0 ^{- 12}})

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

ln (\frac{16384 x ^{12} + 4}{0.1})

What is 1/4 ln(0-1)-1/4 ln(0+3) ?
The solution to 1/4 ln(0-1)-1/4 ln(0+3) is 1/4 ln(-1/3)