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

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

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

simplify

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

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

Decimal Notation

5.99306 \dots

Solution steps

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

Apply log rule:

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

5 ln (2) = ln (2^{5})

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

Apply log rule:

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

5 ln (3) = ln (3^{5})

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

Apply log rule:

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

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

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

2^{5} \cdot 3^{5} = 7776

= ln (7776) - \frac{1}{2} - 5 ln (1) + 1 - 5 ln (2)

Apply log rule:

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

5 ln (2) = ln (2^{5})

= ln (7776) - \frac{1}{2} - 5 ln (1) + 1 - ln (2^{5})

Apply log rule:

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

ln (7776) - ln (2^{5}) = ln (\frac{7776}{2 ^{5}})

= ln (\frac{7776}{2 ^{5}}) - \frac{1}{2} - 5 ln (1) + 1

\frac{7776}{2 ^{5}} = 243

= ln (243) - \frac{1}{2} - 5 ln (1) + 1

5 ln (1) = 0

= ln (243) - \frac{1}{2} - 0 + 1

Add/Subtract the numbers:

- 0 + 1 = 1

= 1 + ln (243) - \frac{1}{2}

Combine the fractions

1 - \frac{1}{2} : \frac{1}{2}

= ln (243) + \frac{1}{2}

ln (243) = 5 ln (3)

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

lo g_{10} (3 x y^{2})

- lo g_{10} (3.8 \cdot 1 0^{- 6})

ln (\frac{7 + e ^{x}}{7 - e ^{x}})

\frac{1}{2} ln (\frac{1}{4})

20 \cdot lo g_{10} (4 \cdot 3.1415 \cdot x \cdot (8.16))

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