What is simplify-1/4 ln(1+2)+1/4 ln(1-2) ?

The solution to simplify-1/4 ln(1+2)+1/4 ln(1-2) is 1/4 ln(-1/3)

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simplify-1/4 ln(1+2)+1/4 ln(1-2)

simplify

- \frac{1}{4} ln (1 + 2) + \frac{1}{4} ln (1 - 2)

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

Solution steps

- \frac{1}{4} ln (1 + 2) + \frac{1}{4} ln (1 - 2)

Apply log rule:

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

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

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

Apply log rule:

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

\frac{1}{4} ln (1 - 2) = ln ((1 - 2)^{\frac{1}{4}})

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

Apply log rule:

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

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

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

3 ln (\frac{3 L}{4})

5.23 + lo g_{10} (\frac{1.76 \cdot 1 0 ^{- 3}}{0.00322})

ln (\frac{25}{75})

4 ln (2) + 2 ln (x) - \frac{1}{2} ln (x + 2)

(\frac{20.5 \cdot 33.2}{12.8}) ln (12.8 \cdot 1 0^{- 3})

What is simplify-1/4 ln(1+2)+1/4 ln(1-2) ?
The solution to simplify-1/4 ln(1+2)+1/4 ln(1-2) is 1/4 ln(-1/3)