What is log_{2}((4*10^{-8})/(128*10^4)) ?

The solution to log_{2}((4*10^{-8})/(128*10^4)) is -12log_{2}(10)-5

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simplify log_{2}((410^{-8})/(12810^4))

simplify

lo g_{2} (\frac{4 \cdot 1 0 ^{- 8}}{128 \cdot 1 0 ^{4}})

- 12 lo g_{2} (10) - 5

Decimal Notation

- 44.86313 \dots

Solution steps

lo g_{2} (\frac{4 \cdot 1 0 ^{- 8}}{128 \cdot 1 0 ^{4}})

Apply log rule:

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

lo g_{2} (\frac{4 \cdot 1 0 ^{- 8}}{128 \cdot 1 0 ^{4}}) = lo g_{2} (4 \cdot 1 0^{- 8}) - lo g_{2} (128 \cdot 1 0^{4})

= lo g_{2} (1 0^{- 8} \cdot 4) - lo g_{2} (1 0^{4} \cdot 128)

Apply log rule:

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

lo g_{2} (4 \cdot 1 0^{- 8}) = lo g_{2} (4) + lo g_{2} (1 0^{- 8}), lo g_{2} (128 \cdot 1 0^{4}) = lo g_{2} (128) + lo g_{2} (1 0^{4})

= lo g_{2} (4) + lo g_{2} (1 0^{- 8}) - (lo g_{2} (128) + lo g_{2} (1 0^{4}))

Apply log rule:

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

lo g_{2} (1 0^{- 8}) = - 8 lo g_{2} (10)

= lo g_{2} (4) - 8 lo g_{2} (10) - (lo g_{2} (128) + lo g_{2} (1 0^{4}))

Apply log rule:

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

lo g_{2} (1 0^{4}) = 4 lo g_{2} (10)

= lo g_{2} (4) - 8 lo g_{2} (10) - (lo g_{2} (128) + 4 lo g_{2} (10))

lo g_{2} (128) = 7

= lo g_{2} (4) - 8 lo g_{2} (10) - (4 lo g_{2} (10) + 7)

lo g_{2} (4) = 2

= 2 - 8 lo g_{2} (10) - (4 lo g_{2} (10) + 7)

- (7 + 4 lo g_{2} (10)) : - 7 - 4 lo g_{2} (10)

= 2 - 8 lo g_{2} (10) - 7 - 4 lo g_{2} (10)

Simplify

2 - 8 lo g_{2} (10) - 7 - 4 lo g_{2} (10) : - 12 lo g_{2} (10) - 5

= - 12 lo g_{2} (10) - 5

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What is log_{2}((4*10^{-8})/(128*10^4)) ?
The solution to log_{2}((4*10^{-8})/(128*10^4)) is -12log_{2}(10)-5