What is 1/5 ln(b^2+3)-1/5 ln(b+3) ?

The solution to 1/5 ln(b^2+3)-1/5 ln(b+3) is ln(\sqrt[5]{(b^2+3)/(b+3)})

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simplify 1/5 ln(b^2+3)-1/5 ln(b+3)

simplify

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

Solution steps

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

Apply log rule:

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

\frac{1}{5} ln (b^{2} + 3) = ln ((b^{2} + 3)^{\frac{1}{5}})

= ln ((b^{2} + 3)^{\frac{1}{5}}) - \frac{1}{5} ln (b + 3)

Apply log rule:

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

\frac{1}{5} ln (b + 3) = ln ((b + 3)^{\frac{1}{5}})

= ln ((b^{2} + 3)^{\frac{1}{5}}) - ln ((b + 3)^{\frac{1}{5}})

Apply log rule:

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

ln ((b^{2} + 3)^{\frac{1}{5}}) - ln ((b + 3)^{\frac{1}{5}}) = ln \frac{( b ^{2} + 3 ) ^{\frac{1}{5}}}{( b + 3 ) ^{\frac{1}{5}}}

= ln \frac{( b ^{2} + 3 ) ^{\frac{1}{5}}}{( b + 3 ) ^{\frac{1}{5}}}

Combine same powers :

\frac{ln ( 6.1 ) - ln ( 6 )}{6.1 - 6}

ln ∣ 7 ∣ + ln ∣ 0.5 ∣

\frac{ln ( 2 )}{ln ( 21.3619 ) - ln ( 58.8018 )}

ln (\frac{62 - 25}{100 - 25})

ln (\frac{17}{25})

What is 1/5 ln(b^2+3)-1/5 ln(b+3) ?
The solution to 1/5 ln(b^2+3)-1/5 ln(b+3) is ln(\sqrt[5]{(b^2+3)/(b+3)})