What is ln(2v^2+1)-2ln(v^2+1)-2ln(y) ?

The solution to ln(2v^2+1)-2ln(v^2+1)-2ln(y) is ln((2v^2+1)/(y^2(v^2+1)^2))

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simplify ln(2v^2+1)-2ln(v^2+1)-2ln(y)

simplify

ln (2 v^{2} + 1) - 2 ln (v^{2} + 1) - 2 ln (y)

ln (\frac{2 v ^{2} + 1}{y ^{2} ( v ^{2} + 1 ) ^{2}})

Solution steps

ln (2 v^{2} + 1) - 2 ln (v^{2} + 1) - 2 ln (y)

Apply log rule:

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

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

= ln (2 v^{2} + 1) - ln ((v^{2} + 1)^{2}) - 2 ln (y)

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

= ln \frac{\frac{2 v ^{2} + 1}{( v ^{2} + 1 ) ^{2}}}{y ^{2}}

Apply the fraction rule:

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

= ln (\frac{2 v ^{2} + 1}{y ^{2} ( v ^{2} + 1 ) ^{2}})

ln (80) + ln (20)

ln (y) + ln (\frac{x}{y} + 1)

ln ∣ x ∣ + ln ∣ x - 1 ∣ + ln ∣ x + 1 ∣

ln ∣ b ∣ ln (\frac{b}{a}) + \frac{- ln ( ∣ b ∣ ^{2} ) + ln ( ∣ a ∣ ^{2} )}{2}

0 \cdot lo g_{10} (0.69) + (1 - 0) \cdot lo g_{10} (1 - 0.69)

What is ln(2v^2+1)-2ln(v^2+1)-2ln(y) ?
The solution to ln(2v^2+1)-2ln(v^2+1)-2ln(y) is ln((2v^2+1)/(y^2(v^2+1)^2))