What is 3ln(2)+1/3 ln(8)-1/2 ln(25) ?

The solution to 3ln(2)+1/3 ln(8)-1/2 ln(25) is ln(16/5)

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simplify 3ln(2)+1/3 ln(8)-1/2 ln(25)

simplify

3 ln (2) + \frac{1}{3} ln (8) - \frac{1}{2} ln (25)

ln (\frac{16}{5})

Decimal Notation

1.16315 \dots

Solution steps

3 ln (2) + \frac{1}{3} ln (8) - \frac{1}{2} ln (25)

Apply log rule:

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

\frac{1}{3} ln (8) = ln (8^{\frac{1}{3}})

= 3 ln (2) + ln (8^{\frac{1}{3}}) - \frac{1}{2} ln (25)

Apply log rule:

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

\frac{1}{2} ln (25) = ln (2 5^{\frac{1}{2}})

= 3 ln (2) + ln (8^{\frac{1}{3}}) - ln (2 5^{\frac{1}{2}})

Apply log rule:

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

ln (8^{\frac{1}{3}}) - ln (2 5^{\frac{1}{2}}) = ln (\frac{8 ^{\frac{1}{3}}}{2 5 ^{\frac{1}{2}}})

= 3 ln (2) + ln (\frac{8 ^{\frac{1}{3}}}{2 5 ^{\frac{1}{2}}})

\frac{8 ^{\frac{1}{3}}}{2 5 ^{\frac{1}{2}}} = \frac{2}{5}

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

Apply log rule:

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

3 ln (2) = ln (2^{3})

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

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

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

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

2^{3} \cdot \frac{2}{5} = \frac{16}{5}

= ln (\frac{16}{5})

2 ln (t) + 5 ln (z) - ln (t)

lo g_{10} (\frac{85347 x ^{15274}}{125386})

- lo g_{10} (1.74 \times 1 0^{- 5})

lo g_{9} (6 x^{3} y^{5} z)

lo g_{5} (\frac{x ^{2}}{y ^{2} z ^{3}})

What is 3ln(2)+1/3 ln(8)-1/2 ln(25) ?
The solution to 3ln(2)+1/3 ln(8)-1/2 ln(25) is ln(16/5)