What is 10log_{10}(2/1)+10log_{10}((0.007)/1)+52 ?

The solution to 10log_{10}(2/1)+10log_{10}((0.007)/1)+52 is log_{10}(1024*7^{10})+22

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simplify 10log_{10}(2/1)+10log_{10}((0.007)/1)+52

simplify

10 lo g_{10} (\frac{2}{1}) + 10 lo g_{10} (\frac{0.007}{1}) + 52

lo g_{10} (1024 \cdot 7^{10}) + 22

Decimal Notation

33.46128 \dots

Solution steps

10 lo g_{10} (\frac{2}{1}) + 10 lo g_{10} (\frac{0.007}{1}) + 52

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

10 lo g_{10} (\frac{0.007}{1}) = 10 lo g_{10} (7) - 30

= 10 lo g_{10} (2) + 10 lo g_{10} (7) - 30 + 52

Add the numbers:

- 30 + 52 = 22

= 10 lo g_{10} (2) + 10 lo g_{10} (7) + 22

10 lo g_{10} (2) = lo g_{10} (2^{10})

10 lo g_{10} (7) = lo g_{10} (7^{10})

= lo g_{10} (2^{10}) + lo g_{10} (7^{10}) + 22

lo g_{10} (2^{10}) + lo g_{10} (7^{10}) = lo g_{10} (1024 \cdot 7^{10})

= lo g_{10} (1024 \cdot 7^{10}) + 22

ln (x^{2} y) - ln (y^{2})

- lo g_{10} (1.25 (1 0^{- 4}))

- lo g_{10} ((1.8 \cdot 1 0^{- 5}) (1))

ln (6 e^{m})

- \frac{1}{2} ln (\frac{1}{16}) + ln (\frac{1}{2})

What is 10log_{10}(2/1)+10log_{10}((0.007)/1)+52 ?
The solution to 10log_{10}(2/1)+10log_{10}((0.007)/1)+52 is log_{10}(1024*7^{10})+22