What is 1/4 ln| 3/2-2/1 |-1/4 ln| 3/2+2/1 | ?

The solution to 1/4 ln| 3/2-2/1 |-1/4 ln| 3/2+2/1 | is ln((7^{3/4})/7)

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simplify 1/4 ln| 3/2-2/1 |-1/4 ln| 3/2+2/1 |

simplify

\frac{1}{4} ln \frac{3}{2} - \frac{2}{1} - \frac{1}{4} ln \frac{3}{2} + \frac{2}{1}

ln (\frac{7 ^{\frac{3}{4}}}{7})

Decimal Notation

- 0.48647 \dots

Solution steps

\frac{1}{4} ln \frac{3}{2} - \frac{2}{1} - \frac{1}{4} ln \frac{3}{2} + \frac{2}{1}

Apply log rule:

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

\frac{1}{4} ln \frac{3}{2} - \frac{2}{1} = ln (\frac{3}{2} - \frac{2}{1}^{\frac{1}{4}})

= ln (\frac{3}{2} - \frac{2}{1}^{\frac{1}{4}}) - \frac{1}{4} ln \frac{3}{2} + \frac{2}{1}

Apply log rule:

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

\frac{1}{4} ln \frac{3}{2} + \frac{2}{1} = ln (\frac{3}{2} + \frac{2}{1}^{\frac{1}{4}})

= ln (\frac{3}{2} - \frac{2}{1}^{\frac{1}{4}}) - ln (\frac{3}{2} + \frac{2}{1}^{\frac{1}{4}})

Apply log rule:

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

ln (\frac{3}{2} - \frac{2}{1}^{\frac{1}{4}}) - ln (\frac{3}{2} + \frac{2}{1}^{\frac{1}{4}}) = ln \frac{\frac{3}{2} - \frac{2}{1} ^{\frac{1}{4}}}{\frac{3}{2} + \frac{2}{1} ^{\frac{1}{4}}}

= ln \frac{\frac{3}{2} - \frac{2}{1} ^{\frac{1}{4}}}{\frac{3}{2} + \frac{2}{1} ^{\frac{1}{4}}}

\frac{\frac{3}{2} - \frac{2}{1} ^{\frac{1}{4}}}{\frac{3}{2} + \frac{2}{1} ^{\frac{1}{4}}} = \frac{7 ^{\frac{3}{4}}}{7}

= ln (\frac{7 ^{\frac{3}{4}}}{7})

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What is 1/4 ln| 3/2-2/1 |-1/4 ln| 3/2+2/1 | ?
The solution to 1/4 ln| 3/2-2/1 |-1/4 ln| 3/2+2/1 | is ln((7^{3/4})/7)