What is 9*ln(6)-ln(2!)+7*ln(2)-ln(7!)-3*6 ?

The solution to 9*ln(6)-ln(2!)+7*ln(2)-ln(7!)-3*6 is ln(5038848)+ln(8/315)-18

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify 9ln(6)-ln(2!)+7ln(2)-ln(7!)-3*6

simplify

9 \cdot ln (6) - ln (2!) + 7 \cdot ln (2) - ln (7!) - 3 \cdot 6

ln (5038848) + ln (\frac{8}{315}) - 18

Decimal Notation

- 6.24044 \dots

Solution steps

9 ln (6) - ln (2!) + 7 ln (2) - ln (7!) - 3 \cdot 6

Apply log rule:

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

9 ln (6) = ln (6^{9})

= ln (6^{9}) - ln (2!) + 7 ln (2) - ln (7!) - 3 \cdot 6

Apply log rule:

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

ln (6^{9}) - ln (2!) = ln (\frac{6 ^{9}}{2 !})

= ln (\frac{6 ^{9}}{2 !}) + 7 ln (2) - ln (7!) - 3 \cdot 6

\frac{6 ^{9}}{2 !} = 5038848

= ln (5038848) + 7 ln (2) - ln (7!) - 3 \cdot 6

Apply log rule:

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

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

= ln (5038848) + ln (2^{7}) - ln (7!) - 3 \cdot 6

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 (5038848) + ln (\frac{2 ^{7}}{7 !}) - 3 \cdot 6

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

= ln (5038848) + ln (\frac{8}{315}) - 3 \cdot 6

Multiply the numbers:

3 \cdot 6 = 18

= ln (5038848) + ln (\frac{8}{315}) - 18

ln (\frac{0.69}{0.664})

\frac{5 ln x ^{2} - 9 - ln ∣ x + 3 ∣ + ln ∣ x - 3 ∣ 2}{2} + C

\frac{1}{2} ln (\frac{2240}{1800})

ln (\frac{( e ^{- 2 x} )}{2})

\frac{1}{3} - \frac{3}{2} + 9 + 27 ln (4) - 27 ln (3)

What is 9*ln(6)-ln(2!)+7*ln(2)-ln(7!)-3*6 ?
The solution to 9*ln(6)-ln(2!)+7*ln(2)-ln(7!)-3*6 is ln(5038848)+ln(8/315)-18