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

The solution to 1/3 ln(5+1)-1/3 ln(5-2) is 1/3 ln(2)

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

simplify

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

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

Decimal Notation

0.23104 \dots

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Rewrite as

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

Apply log rule

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

assuming

x \geq 0

= \frac{1}{3} ln (2)

ln (3.17 \cdot 1 0^{- 9})

lo g_{\frac{2}{π}} (2 \cdot π)

lo g_{10} (z^{6} y)

ln (\frac{168.9 2 ^{n}}{146. 2 ^{n}})

20 lo g_{10} (\frac{1}{( i w ) ^{2}})

What is 1/3 ln(5+1)-1/3 ln(5-2) ?
The solution to 1/3 ln(5+1)-1/3 ln(5-2) is 1/3 ln(2)