What is 1/2 ln(u)-5/8 ln(2u^2+1) ?

The solution to 1/2 ln(u)-5/8 ln(2u^2+1) is ln((u^{1/2}(2u^2+1)^{3/8})/(2u^2+1))

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simplify 1/2 ln(u)-5/8 ln(2u^2+1)

simplify

\frac{1}{2} ln (u) - \frac{5}{8} ln (2 u^{2} + 1)

ln \frac{u ^{\frac{1}{2}} ( 2 u ^{2} + 1 ) ^{\frac{3}{8}}}{2 u ^{2} + 1}

Solution steps

\frac{1}{2} ln (u) - \frac{5}{8} ln (2 u^{2} + 1)

Apply log rule:

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

\frac{1}{2} ln (u) = ln (u^{\frac{1}{2}})

= ln (u^{\frac{1}{2}}) - \frac{5}{8} ln (2 u^{2} + 1)

Apply log rule:

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

\frac{5}{8} ln (2 u^{2} + 1) = ln ((2 u^{2} + 1)^{\frac{5}{8}})

= ln (u^{\frac{1}{2}}) - ln ((2 u^{2} + 1)^{\frac{5}{8}})

Apply log rule:

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

ln (u^{\frac{1}{2}}) - ln ((2 u^{2} + 1)^{\frac{5}{8}}) = ln (\frac{u ^{\frac{1}{2}}}{( 2 u ^{2} + 1 ) ^{\frac{5}{8}}})

= ln (\frac{u ^{\frac{1}{2}}}{( 2 u ^{2} + 1 ) ^{\frac{5}{8}}})

Simplify

\frac{u ^{\frac{1}{2}}}{( 2 u ^{2} + 1 ) ^{\frac{5}{8}}} : \frac{u ^{\frac{1}{2}} ( 2 u ^{2} + 1 ) ^{\frac{3}{8}}}{2 u ^{2} + 1}

= ln \frac{u ^{\frac{1}{2}} ( 2 u ^{2} + 1 ) ^{\frac{3}{8}}}{2 u ^{2} + 1}

ln (\frac{24.8}{24.5})

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ln (3.436 - 1) - ln (1.718)

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ln (2) - \frac{1}{3} ln (5)

What is 1/2 ln(u)-5/8 ln(2u^2+1) ?
The solution to 1/2 ln(u)-5/8 ln(2u^2+1) is ln((u^{1/2}(2u^2+1)^{3/8})/(2u^2+1))