What is 1/2 ln(5+2)-1/2 ln(5+0) ?

The solution to 1/2 ln(5+2)-1/2 ln(5+0) is 1/2 ln(7/5)

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simplify 1/2 ln(5+2)-1/2 ln(5+0)

simplify

\frac{1}{2} ln (5 + 2) - \frac{1}{2} ln (5 + 0)

\frac{1}{2} ln (\frac{7}{5})

Decimal Notation

0.16823 \dots

Solution steps

\frac{1}{2} ln (5 + 2) - \frac{1}{2} ln (5 + 0)

Apply log rule:

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

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

= ln ((5 + 2)^{\frac{1}{2}}) - \frac{1}{2} ln (5 + 0)

Apply log rule:

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

\frac{1}{2} ln (5 + 0) = ln ((5 + 0)^{\frac{1}{2}})

= ln ((5 + 2)^{\frac{1}{2}}) - ln ((5 + 0)^{\frac{1}{2}})

Apply log rule:

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

ln ((5 + 2)^{\frac{1}{2}}) - ln ((5 + 0)^{\frac{1}{2}}) = ln (\frac{( 5 + 2 ) ^{\frac{1}{2}}}{( 5 + 0 ) ^{\frac{1}{2}}})

= ln (\frac{( 5 + 2 ) ^{\frac{1}{2}}}{( 5 + 0 ) ^{\frac{1}{2}}})

\frac{( 5 + 2 ) ^{\frac{1}{2}}}{( 5 + 0 ) ^{\frac{1}{2}}} = \frac{7}{5}

= ln (\frac{7}{5})

Rewrite as

= ln ((\frac{7}{5})^{\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{7}{5})

ln (\frac{1}{( 3.5 e ^{(- 4 T)} - 5 e ^{(- T)} + 2.5 )})

lo g_{b} (\frac{a ^{2} 2}{e ^{3}})

ln (v^{- \frac{1}{3}}) + ln ((v - 3)^{\frac{1}{3}}) + c

13 lo g_{10} (a) - lo g_{10} (b)

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

What is 1/2 ln(5+2)-1/2 ln(5+0) ?
The solution to 1/2 ln(5+2)-1/2 ln(5+0) is 1/2 ln(7/5)