What is ln(1^2(1+2)^{9/5}(1-3)^{1/5}) ?

The solution to ln(1^2(1+2)^{9/5}(1-3)^{1/5}) is (9ln(3)+ln(-2))/5

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simplify ln(1^2(1+2)^{9/5}(1-3)^{1/5})

simplify

ln (1^{2} (1 + 2)^{\frac{9}{5}} (1 - 3)^{\frac{1}{5}})

\frac{9 ln ( 3 ) + ln ( - 2 )}{5}

Solution steps

ln (1^{2} (1 + 2)^{\frac{9}{5}} (1 - 3)^{\frac{1}{5}})

Apply log rule:

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

ln (1^{2} (1 + 2)^{\frac{9}{5}} (1 - 3)^{\frac{1}{5}}) = ln (1^{2}) + ln ((1 + 2)^{\frac{9}{5}}) + ln ((1 - 3)^{\frac{1}{5}})

= ln (1^{2}) + ln ((1 + 2)^{\frac{9}{5}}) + ln ((1 - 3)^{\frac{1}{5}})

Apply log rule:

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

ln (1^{2}) = 2 ln (1)

= 2 ln (1) + ln ((1 + 2)^{\frac{9}{5}}) + ln ((1 - 3)^{\frac{1}{5}})

Apply log rule:

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

ln ((1 + 2)^{\frac{9}{5}}) = \frac{9}{5} ln (1 + 2)

= 2 ln (1) + \frac{9}{5} ln (1 + 2) + ln ((1 - 3)^{\frac{1}{5}})

Apply log rule:

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

ln ((1 - 3)^{\frac{1}{5}}) = \frac{1}{5} ln (1 - 3)

= 2 ln (1) + \frac{9}{5} ln (1 + 2) + \frac{1}{5} ln (1 - 3)

2 ln (1) = 0

\frac{9}{5} ln (1 + 2) = \frac{9 ln ( 3 )}{5}

\frac{1}{5} ln (1 - 3) = \frac{ln ( - 2 )}{5}

= 0 + \frac{9 ln ( 3 )}{5} + \frac{ln ( - 2 )}{5}

Combine the fractions

\frac{9 ln ( 3 )}{5} + \frac{ln ( - 2 )}{5} : \frac{9 ln ( 3 ) + ln ( - 2 )}{5}

= 0 + \frac{9 ln ( 3 ) + ln ( - 2 )}{5}

0 + \frac{9 ln ( 3 ) + ln ( - 2 )}{5} = \frac{9 ln ( 3 ) + ln ( - 2 )}{5}

= \frac{9 ln ( 3 ) + ln ( - 2 )}{5}

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

\frac{3}{5} lo g_{\frac{4}{25}} (5) - \frac{3}{5} lo g_{\frac{4}{25}} (2)

\frac{3}{4} \cdot lo g_{2} (\frac{( \frac{3}{4} )}{( \frac{1}{20} )})

\frac{8}{π} \cdot ln (6) + \frac{2}{π} \cdot ln (\frac{7}{6})

ln (\frac{( x - 3 )}{( x + 3 )})

What is ln(1^2(1+2)^{9/5}(1-3)^{1/5}) ?
The solution to ln(1^2(1+2)^{9/5}(1-3)^{1/5}) is (9ln(3)+ln(-2))/5