What is ln((2.3113*10^{-1})/(4.98*10^{-3)}) ?

The solution to ln((2.3113*10^{-1})/(4.98*10^{-3)}) is 2ln(10)-ln(4.98)+ln(2.3113)

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simplify ln((2.311310^{-1})/(4.9810^{-3)})

simplify

ln (\frac{2.3113 \cdot 1 0 ^{- 1}}{4.98 \cdot 1 0 ^{- 3}})

2 ln (10) - ln (4.98) + ln (2.3113)

Decimal Notation

3.83755 \dots

Solution steps

ln (\frac{2.3113 \cdot 1 0 ^{- 1}}{4.98 \cdot 1 0 ^{- 3}})

Apply log rule:

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

ln (\frac{2.3113 \cdot 1 0 ^{- 1}}{4.98 \cdot 1 0 ^{- 3}}) = ln (2.3113 \cdot 1 0^{- 1}) - ln (4.98 \cdot 1 0^{- 3})

= ln (1 0^{- 1} \cdot 2.3113) - ln (1 0^{- 3} \cdot 4.98)

Apply log rule:

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

ln (2.3113 \cdot 1 0^{- 1}) = ln (2.3113) + ln (1 0^{- 1}), ln (4.98 \cdot 1 0^{- 3}) = ln (4.98) + ln (1 0^{- 3})

= ln (2.3113) + ln (1 0^{- 1}) - (ln (4.98) + ln (1 0^{- 3}))

Apply log rule:

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

ln (1 0^{- 1}) = - 1 \cdot ln (10)

= ln (2.3113) - 1 \cdot ln (10) - (ln (4.98) + 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 (2.3113) - 1 \cdot ln (10) - (ln (4.98) - 3 ln (10))

Multiply:

1 \cdot ln (10) = ln (10)

= ln (2.3113) - ln (10) - (ln (4.98) - 3 ln (10))

- (ln (4.98) - 3 ln (10)) : - ln (4.98) + 3 ln (10)

= ln (2.3113) - ln (10) - ln (4.98) + 3 ln (10)

Simplify

ln (2.3113) - ln (10) - ln (4.98) + 3 ln (10) : 2 ln (10) - ln (4.98) + ln (2.3113)

= 2 ln (10) - ln (4.98) + ln (2.3113)

\frac{1}{4} ln \frac{3}{2} - \frac{2}{1} - \frac{1}{4} ln \frac{3}{2} + \frac{2}{1}

lo g_{2} (x - 5) + lo g_{2} (x - 11)

2 lo g_{3} (x) - 3 lo g_{3} (y) + 5 lo g_{3} (z)

\frac{1}{2} lo g_{e} (1 + \frac{2 y}{x}) + lo g_{e} (x)

\frac{1}{2} lo g_{10} (7)

What is ln((2.3113*10^{-1})/(4.98*10^{-3)}) ?
The solution to ln((2.3113*10^{-1})/(4.98*10^{-3)}) is 2ln(10)-ln(4.98)+ln(2.3113)