What is 15ln(5)-24ln(4)+9ln(3)-2ln(2) ?

The solution to 15ln(5)-24ln(4)+9ln(3)-2ln(2) is ln((5^{15}*19683)/(2^{50)})

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simplify 15ln(5)-24ln(4)+9ln(3)-2ln(2)

simplify

15 ln (5) - 24 ln (4) + 9 ln (3) - 2 ln (2)

ln (\frac{5 ^{15} \cdot 19683}{2 ^{50}})

Decimal

- 0.62827 \dots

Solution steps

15 ln (5) - 24 ln (4) + 9 ln (3) - 2 ln (2)

Apply log rule:

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

15 ln (5) = ln (5^{15})

= ln (5^{15}) - 24 ln (4) + 9 ln (3) - 2 ln (2)

Apply log rule:

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

24 ln (4) = ln (4^{24})

= ln (5^{15}) - ln (4^{24}) + 9 ln (3) - 2 ln (2)

Apply log rule:

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

ln (5^{15}) - ln (4^{24}) = ln (\frac{5 ^{15}}{4 ^{24}})

= ln (\frac{5 ^{15}}{4 ^{24}}) + 9 ln (3) - 2 ln (2)

Apply log rule:

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

9 ln (3) = ln (3^{9})

= ln (\frac{5 ^{15}}{4 ^{24}}) + ln (3^{9}) - 2 ln (2)

Apply log rule:

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

ln (\frac{5 ^{15}}{4 ^{24}}) + ln (3^{9}) = ln (3^{9} \cdot \frac{5 ^{15}}{4 ^{24}})

= ln (3^{9} \cdot \frac{5 ^{15}}{4 ^{24}}) - 2 ln (2)

\frac{5 ^{15}}{4 ^{24}} \cdot 3^{9} = \frac{5 ^{15} \cdot 19683}{4 ^{24}}

= ln (\frac{5 ^{15} \cdot 19683}{4 ^{24}}) - 2 ln (2)

Apply log rule:

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

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

= ln (\frac{5 ^{15} \cdot 19683}{4 ^{24}}) - ln (2^{2})

Apply log rule:

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

ln (\frac{5 ^{15} \cdot 19683}{4 ^{24}}) - ln (2^{2}) = ln (\frac{\frac{5 ^{15} \cdot 19683}{4 ^{24}}}{2 ^{2}})

= ln (\frac{\frac{5 ^{15} \cdot 19683}{4 ^{24}}}{2 ^{2}})

\frac{\frac{5 ^{15} \cdot 19683}{4 ^{24}}}{2 ^{2}} = \frac{5 ^{15} \cdot 19683}{2 ^{50}}

= ln (\frac{5 ^{15} \cdot 19683}{2 ^{50}})

\frac{1}{10} ln (\frac{1}{3})

ln (5 e^{- \frac{6 x}{t}})

- lo g_{10} (5.4 \cdot 1 0^{- 6})

lo g_{e} (\frac{56}{77})

ln (\frac{5 x ^{3} y ^{2}}{z ^{4}})

What is 15ln(5)-24ln(4)+9ln(3)-2ln(2) ?
The solution to 15ln(5)-24ln(4)+9ln(3)-2ln(2) is ln((5^{15}*19683)/(2^{50)})