What is log_{10}(1300/2895)+log_{10}(0.965) ?

The solution to log_{10}(1300/2895)+log_{10}(0.965) is log_{10}(13/3)-1

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simplify log_{10}(1300/2895)+log_{10}(0.965)

simplify

lo g_{10} (\frac{1300}{2895}) + lo g_{10} (0.965)

lo g_{10} (\frac{13}{3}) - 1

Decimal Notation

- 0.36317 \dots

Solution steps

lo g_{10} (\frac{1300}{2895}) + lo g_{10} (0.965)

lo g_{10} (\frac{1300}{2895}) = lo g_{10} (\frac{260}{579})

lo g_{10} (0.965) = lo g_{10} (\frac{193}{200})

= lo g_{10} (\frac{260}{579}) + lo g_{10} (\frac{193}{200})

Apply log rule:

lo g_{a} (x) + lo g_{a} (y) = lo g_{a} (x y)

lo g_{10} (\frac{260}{579}) + lo g_{10} (\frac{193}{200}) = lo g_{10} (\frac{260}{579} \cdot \frac{193}{200})

= lo g_{10} (\frac{260}{579} \cdot \frac{193}{200})

\frac{260}{579} \cdot \frac{193}{200} = \frac{13}{30}

= lo g_{10} (\frac{13}{30})

Apply log rule:

lo g_{a} (\frac{x}{y}) = lo g_{a} (x) - lo g_{a} (y)

lo g_{10} (\frac{13}{30}) = lo g_{10} (13) - lo g_{10} (30)

= lo g_{10} (13) - lo g_{10} (30)

lo g_{10} (30) = 1 + lo g_{10} (3)

= lo g_{10} (13) - (1 + lo g_{10} (3))

Apply the distributive law:

- (a + b) = - a - b

- (1 + lo g_{10} (3)) = - 1 - lo g_{10} (3)

= lo g_{10} (13) - 1 - lo g_{10} (3)

Apply log rule:

lo g_{a} (x) - lo g_{a} (y) = lo g_{a} (\frac{x}{y})

lo g_{10} (13) - lo g_{10} (3) = lo g_{10} (\frac{13}{3})

= lo g_{10} (\frac{13}{3}) - 1

2 ln ∣ x ∣ - ln x^{2} + 1 + arctan (x) + \frac{1}{x ^{2} + 1}

ln (55)

ln (x) - 3 ln (y) - ln (4)

ln (6 + x^{2}) - ln (3 + x)

5.35 \cdot ln (\frac{693}{298}) + 2.4 \cdot 1 0^{- 3} \cdot (693 - 298)

What is log_{10}(1300/2895)+log_{10}(0.965) ?
The solution to log_{10}(1300/2895)+log_{10}(0.965) is log_{10}(13/3)-1