What is ln(0.1^2*0.5^2*1^2*4*16) ?

The solution to ln(0.1^2*0.5^2*1^2*4*16) is 2ln(0.1)+2ln(0.5)+6ln(2)

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simplify ln(0.1^20.5^21^2416)

simplify

ln (0. 1^{2} \cdot 0. 5^{2} \cdot 1^{2} \cdot 4 \cdot 16)

2 ln (0.1) + 2 ln (0.5) + 6 ln (2)

Decimal Notation

- 1.83258 \dots

Solution steps

ln (0. 1^{2} \cdot 0. 5^{2} \cdot 1^{2} \cdot 4 \cdot 16)

Apply log rule:

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

ln (0. 1^{2} \cdot 0. 5^{2} \cdot 1^{2} \cdot 4 \cdot 16) = ln (0. 1^{2}) + ln (0. 5^{2}) + ln (1^{2}) + ln (4) + ln (16)

= ln (0. 1^{2}) + ln (0. 5^{2}) + ln (1^{2}) + ln (4) + ln (16)

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (0. 1^{2}) = 2 ln (0.1)

= 2 ln (0.1) + ln (0. 5^{2}) + ln (1^{2}) + ln (4) + ln (16)

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (0. 5^{2}) = 2 ln (0.5)

= 2 ln (0.1) + 2 ln (0.5) + ln (1^{2}) + ln (4) + ln (16)

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

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

= 2 ln (0.1) + 2 ln (0.5) + 2 ln (1) + ln (4) + ln (16)

2 ln (1) = 0

= 2 ln (0.1) + 2 ln (0.5) + 0 + ln (4) + ln (16)

2 ln (0.1) + 2 ln (0.5) + 0 + ln (4) + ln (16) = 2 ln (0.1) + 2 ln (0.5) + ln (4) + ln (16)

= 2 ln (0.1) + 2 ln (0.5) + ln (4) + ln (16)

ln (4) = 2 ln (2)

ln (16) = 4 ln (2)

= 2 ln (0.1) + 2 ln (0.5) + 2 ln (2) + 4 ln (2)

Add similar elements:

2 ln (2) + 4 ln (2) = 6 ln (2)

= 2 ln (0.1) + 2 ln (0.5) + 6 ln (2)

ln (\frac{85}{75})

ln (\frac{x ^{2} + 2 x - 24}{x})

4 lo g_{9} (u) + 24 lo g_{9} (v) + 4 lo g_{9} (w)

ln (\frac{0.023}{0.015})

- lo g_{10} (7.1 \cdot 1 0^{- 3})

What is ln(0.1^2*0.5^2*1^2*4*16) ?
The solution to ln(0.1^2*0.5^2*1^2*4*16) is 2ln(0.1)+2ln(0.5)+6ln(2)