What is 2ln(10)-ln(5)-2ln(2)+ln(3) ?

The solution to 2ln(10)-ln(5)-2ln(2)+ln(3) is ln(15)

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simplify 2ln(10)-ln(5)-2ln(2)+ln(3)

simplify

2 ln (10) - ln (5) - 2 ln (2) + ln (3)

ln (15)

Decimal Notation

2.70805 \dots

Solution steps

2 ln (10) - ln (5) - 2 ln (2) + ln (3)

Apply log rule:

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

2 ln (10) = ln (1 0^{2})

= ln (1 0^{2}) - ln (5) - 2 ln (2) + ln (3)

Apply log rule:

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

ln (1 0^{2}) - ln (5) = ln (\frac{1 0 ^{2}}{5})

= ln (\frac{1 0 ^{2}}{5}) - 2 ln (2) + ln (3)

\frac{1 0 ^{2}}{5} = 20

= ln (20) - 2 ln (2) + ln (3)

Apply log rule:

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

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

= ln (20) - ln (2^{2}) + ln (3)

Apply log rule:

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

ln (20) - ln (2^{2}) = ln (\frac{20}{2 ^{2}})

= ln (\frac{20}{2 ^{2}}) + ln (3)

\frac{20}{2 ^{2}} = 5

= ln (5) + ln (3)

Apply log rule:

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

ln (5) + ln (3) = ln (5 \cdot 3)

= ln (5 \cdot 3)

Multiply the numbers:

5 \cdot 3 = 15

= ln (15)

- lo g_{10} (8.41 \cdot 0.001)

ln (\frac{0.5}{1.5})

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

ln (\frac{r}{2 s})

ln (22.4 \cdot 1 0^{- 7})

What is 2ln(10)-ln(5)-2ln(2)+ln(3) ?
The solution to 2ln(10)-ln(5)-2ln(2)+ln(3) is ln(15)