What is ln((95*10^{12}(2.5))/(10^{-3)}) ?

The solution to ln((95*10^{12}(2.5))/(10^{-3)}) is ln(95)+12ln(10)+ln(2.5)--3ln(10)

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simplify ln((95*10^{12}(2.5))/(10^{-3)})

simplify

ln (\frac{95 \cdot 1 0 ^{12} ( 2.5 )}{1 0 ^{- 3}})

ln (95) + 12 ln (10) + ln (2.5) - - 3 ln (10)

Decimal Notation

26.19343 \dots

Solution steps

ln (\frac{95 \cdot 1 0 ^{12} ( 2.5 )}{1 0 ^{- 3}})

Apply log rule:

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

ln (\frac{95 \cdot 1 0 ^{12} ( 2.5 )}{1 0 ^{- 3}}) = ln (95 \cdot 1 0^{12} (2.5)) - ln (1 0^{- 3})

= ln (1 0^{12} \cdot 95 \cdot 2.5) - ln (1 0^{- 3})

Apply log rule:

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

ln (1 0^{- 3}) = - 3 ln (10)

= ln (95 \cdot 1 0^{12} \cdot 2.5) - - 3 ln (10)

Apply log rule:

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

ln (95 \cdot 1 0^{12} \cdot 2.5) = ln (95) + ln (1 0^{12}) + ln (2.5)

= ln (95) + ln (1 0^{12}) + ln (2.5) - - 3 ln (10)

Apply log rule:

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

ln (1 0^{12}) = 12 ln (10)

= ln (95) + 12 ln (10) + ln (2.5) - - 3 ln (10)

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

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

ln (\frac{1643.85}{547.85}) \cdot 10

\frac{1}{2} ln (\frac{u - a}{u + a})

ln (\frac{3 x}{7})

What is ln((95*10^{12}(2.5))/(10^{-3)}) ?
The solution to ln((95*10^{12}(2.5))/(10^{-3)}) is ln(95)+12ln(10)+ln(2.5)--3ln(10)