What is 20log_{10}(4pi)-20log_{10}(300000) ?

The solution to 20log_{10}(4pi)-20log_{10}(300000) is log_{10}((pi^{20})/(5^{100)*2^{60}*3^{20}})

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simplify 20log_{10}(4pi)-20log_{10}(300000)

simplify

20 lo g_{10} (4 π) - 20 lo g_{10} (300000)

lo g_{10} (\frac{π ^{20}}{5 ^{100} \cdot 2 ^{60} \cdot 3 ^{20}})

Decimal Notation

- 87.55822 \dots

Solution steps

20 lo g_{10} (4 π) - 20 lo g_{10} (300000)

Apply log rule:

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

20 lo g_{10} (4 π) = lo g_{10} ((4 π)^{20})

= lo g_{10} ((4 π)^{20}) - 20 lo g_{10} (300000)

Apply log rule:

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

20 lo g_{10} (300000) = lo g_{10} (30000 0^{20})

= lo g_{10} ((4 π)^{20}) - lo g_{10} (30000 0^{20})

Apply log rule:

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

lo g_{10} ((4 π)^{20}) - lo g_{10} (30000 0^{20}) = lo g_{10} (\frac{( 4 π ) ^{20}}{30000 0 ^{20}})

= lo g_{10} (\frac{( 4 π ) ^{20}}{30000 0 ^{20}})

\frac{( 4 π ) ^{20}}{30000 0 ^{20}} = \frac{π ^{20}}{5 ^{100} \cdot 2 ^{60} \cdot 3 ^{20}}

= lo g_{10} (\frac{π ^{20}}{5 ^{100} \cdot 2 ^{60} \cdot 3 ^{20}})

- ln ∣ x + 2 ∣ + 2 ln ∣ x - 3 ∣ + C

1 \cdot lo g_{10} (0.65) + (1 - 1) \cdot lo g_{10} (1 - 0.65)

ln (x) + ln (y) + 2 ln (12 - x - y)

ln (\frac{x}{( x + 1 )})

\frac{2}{3} lo g_{2} (x) + lo g_{2} (y)

What is 20log_{10}(4pi)-20log_{10}(300000) ?
The solution to 20log_{10}(4pi)-20log_{10}(300000) is log_{10}((pi^{20})/(5^{100)*2^{60}*3^{20}})