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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Combine same powers :

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

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

4 lo g_{10} (2) + lo g_{10}^{2} (5)

- lo g_{10} (6 \cdot 1 0^{- 10})

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

- 6 (ln (3) - ln (2)) + \frac{7}{2} - 6 ln (2) + \frac{15}{2}

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