What is ln(5+x^2)+1/9 ln(x)-ln(x+1) ?

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

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simplify ln(5+x^2)+1/9 ln(x)-ln(x+1)

simplify

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

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

Solution steps

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

Apply log rule:

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

\frac{1}{9} ln (x) = ln (x^{\frac{1}{9}})

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

lo g_{4} (x) + 6 lo g_{4} (y) + 3 lo g_{4} (z)

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

lo g_{6} (\frac{6}{x})

- 100 ln (\frac{1}{100})

ln (\frac{1}{1 - 0.6})

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