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

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

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

simplify

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

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

Decimal Notation

0.5

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Combine same powers :

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

Rewrite as

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

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

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

ln (\frac{l}{2})

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