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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

ln ((x^{2} + 1)^{5}) = 5 ln (x^{2} + 1)

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

Apply log rule:

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

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

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

6 - lo g_{10} (\frac{0.035}{0.1})

- lo g_{10} (7.57 \cdot 1 0^{- 3})

\frac{1}{0.81 \cdot 1 0 ^{- 4}} ln (\frac{1.04}{0.03})

ln (\frac{( s - a )}{s - b})

ln (A B^{n})

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