What is 9*ln(3)-ln(2!)+7*ln(2)-ln(7!)-3*3 ?

The solution to 9*ln(3)-ln(2!)+7*ln(2)-ln(7!)-3*3 is ln(8748/35)-9

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simplify 9ln(3)-ln(2!)+7ln(2)-ln(7!)-3*3

simplify

9 \cdot ln (3) - ln (2!) + 7 \cdot ln (2) - ln (7!) - 3 \cdot 3

ln (\frac{8748}{35}) - 9

Decimal Notation

- 3.47876 \dots

Solution steps

9 ln (3) - ln (2!) + 7 ln (2) - ln (7!) - 3 \cdot 3

Apply log rule:

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

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

= ln (3^{9}) - ln (2!) + 7 ln (2) - ln (7!) - 3 \cdot 3

Apply log rule:

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

ln (3^{9}) - ln (2!) = ln (\frac{3 ^{9}}{2 !})

= ln (\frac{3 ^{9}}{2 !}) + 7 ln (2) - ln (7!) - 3 \cdot 3

\frac{3 ^{9}}{2 !} = \frac{19683}{2}

= ln (\frac{19683}{2}) + 7 ln (2) - ln (7!) - 3 \cdot 3

Apply log rule:

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

7 ln (2) = ln (2^{7})

= ln (\frac{19683}{2}) + ln (2^{7}) - ln (7!) - 3 \cdot 3

Apply log rule:

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

ln (2^{7}) - ln (7!) = ln (\frac{2 ^{7}}{7 !})

= ln (\frac{19683}{2}) + ln (\frac{2 ^{7}}{7 !}) - 3 \cdot 3

\frac{2 ^{7}}{7 !} = \frac{8}{315}

= ln (\frac{19683}{2}) + ln (\frac{8}{315}) - 3 \cdot 3

Apply log rule:

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

ln (\frac{19683}{2}) + ln (\frac{8}{315}) = ln (\frac{8}{315} \cdot \frac{19683}{2})

= ln (\frac{8}{315} \cdot \frac{19683}{2}) - 3 \cdot 3

\frac{19683}{2} \cdot \frac{8}{315} = \frac{8748}{35}

= ln (\frac{8748}{35}) - 3 \cdot 3

Multiply the numbers:

3 \cdot 3 = 9

= ln (\frac{8748}{35}) - 9

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What is 9*ln(3)-ln(2!)+7*ln(2)-ln(7!)-3*3 ?
The solution to 9*ln(3)-ln(2!)+7*ln(2)-ln(7!)-3*3 is ln(8748/35)-9