What is 20*log_{10}(1/(2pi^0+1)) ?

The solution to 20*log_{10}(1/(2pi^0+1)) is -20log_{10}(3)

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simplify 20*log_{10}(1/(2pi^0+1))

simplify

20 \cdot lo g_{10} (\frac{1}{2 π ^{0} + 1})

- 20 lo g_{10} (3)

Decimal Notation

- 9.54242 \dots

Solution steps

20 lo g_{10} (\frac{1}{2 π ^{0} + 1})

Apply log rule:

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

lo g_{10} (\frac{1}{2 π ^{0} + 1}) = lo g_{10} (1) - lo g_{10} (2 π^{0} + 1)

= 20 (lo g_{10} (1) - lo g_{10} (2 π^{0} + 1))

lo g_{10} (1) = 0

lo g_{10} (2 π^{0} + 1) = lo g_{10} (3)

= 20 (- lo g_{10} (3) + 0)

0 - lo g_{10} (3) = - lo g_{10} (3)

= 20 (- lo g_{10} (3))

Remove parentheses:

(- a) = - a

= - 20 lo g_{10} (3)

- (\frac{5}{11}) \cdot ln (\frac{5}{11})

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3 ln (\frac{13}{x})

ln (\frac{26}{\frac{7}{0.05}})

What is 20*log_{10}(1/(2pi^0+1)) ?
The solution to 20*log_{10}(1/(2pi^0+1)) is -20log_{10}(3)