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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

3 lo g_{2} (\frac{5 x}{7})

- lo g_{10} (3.6 e^{- 13})

ln (7.83) - 6 ln (10)

\frac{ln ( 36731 ) - ln ( 27682 )}{2021 - 2000}

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