What is 0.491*log_{10}(21)-0.491*log_{10}(20) ?

The solution to 0.491*log_{10}(21)-0.491*log_{10}(20) is 0.491log_{10}(21)-0.491-0.491log_{10}(2)

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simplify 0.491log_{10}(21)-0.491log_{10}(20)

simplify

0.491 \cdot lo g_{10} (21) - 0.491 \cdot lo g_{10} (20)

0.491 lo g_{10} (21) - 0.491 - 0.491 lo g_{10} (2)

Decimal Notation

0.01040 \dots

Solution steps

0.491 lo g_{10} (21) - 0.491 lo g_{10} (20)

0.491 lo g_{10} (20) = 0.491 + 0.491 lo g_{10} (2)

= 0.491 lo g_{10} (21) - (0.491 + 0.491 lo g_{10} (2))

Apply the distributive law:

- (a + b) = - a - b

- (0.491 + 0.491 lo g_{10} (2)) = - 0.491 - 0.491 lo g_{10} (2)

= 0.491 lo g_{10} (21) - 0.491 - 0.491 lo g_{10} (2)

lo g_{x} (x^{2} y)

- \frac{3}{6} lo g_{10} (\frac{3}{6}) - \frac{3}{6} lo g_{10} (\frac{3}{6})

ln (\frac{e ^{2}}{9})

lo g_{10} (3 x) - lo g_{10} (x)

ln (x e^{b x})

What is 0.491*log_{10}(21)-0.491*log_{10}(20) ?
The solution to 0.491*log_{10}(21)-0.491*log_{10}(20) is 0.491log_{10}(21)-0.491-0.491log_{10}(2)