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

The solution to ln(m)-1/2 ln(m^2+1) is ln((msqrt(m^2+1))/(m^2+1))

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simplify ln(m)-1/2 ln(m^2+1)

simplify

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

ln (\frac{m m ^{2} + 1}{m ^{2} + 1})

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Simplify

\frac{m}{( m ^{2} + 1 ) ^{\frac{1}{2}}} : \frac{m m ^{2} + 1}{m ^{2} + 1}

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

lo g_{y} (\frac{z}{x})

ln (\frac{x}{x ^{2}})

ln (x + 6) - ln (7)

- lo g_{10} (1.8 \cdot 1 0^{- 4} (\frac{0.025}{0.0375}))

lo g_{10} (5 \cdot 2) - lo g_{10} (5) \cdot lo g_{10} (2)

What is ln(m)-1/2 ln(m^2+1) ?
The solution to ln(m)-1/2 ln(m^2+1) is ln((msqrt(m^2+1))/(m^2+1))