What is 2log_{10}(12)-3log_{10}(6)+2log_{10}(2) ?

The solution to 2log_{10}(12)-3log_{10}(6)+2log_{10}(2) is log_{10}(8/3)

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simplify 2log_{10}(12)-3log_{10}(6)+2log_{10}(2)

simplify

2 lo g_{10} (12) - 3 lo g_{10} (6) + 2 lo g_{10} (2)

lo g_{10} (\frac{8}{3})

Decimal Notation

0.42596 \dots

Solution steps

2 lo g_{10} (12) - 3 lo g_{10} (6) + 2 lo g_{10} (2)

2 lo g_{10} (12) = lo g_{10} (1 2^{2})

- 3 lo g_{10} (6) = - lo g_{10} (6^{3})

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

= lo g_{10} (1 2^{2}) - lo g_{10} (6^{3}) + lo g_{10} (2^{2})

Apply log rule:

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

lo g_{10} (1 2^{2}) - lo g_{10} (6^{3}) = lo g_{10} (\frac{1 2 ^{2}}{6 ^{3}})

= lo g_{10} (\frac{1 2 ^{2}}{6 ^{3}}) + lo g_{10} (2^{2})

Apply log rule:

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

lo g_{10} (\frac{1 2 ^{2}}{6 ^{3}}) + lo g_{10} (2^{2}) = lo g_{10} (\frac{1 2 ^{2}}{6 ^{3}} \cdot 2^{2})

= lo g_{10} (\frac{1 2 ^{2}}{6 ^{3}} \cdot 2^{2})

\frac{1 2 ^{2}}{6 ^{3}} \cdot 2^{2} = \frac{8}{3}

= lo g_{10} (\frac{8}{3})

lo g_{7} ((\frac{x ^{5}}{y})^{5})

ln (x ln (x))

lo g_{10} (7) - lo g_{10} (x)

lo g_{10} (2) + lo g_{10} (p)

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

What is 2log_{10}(12)-3log_{10}(6)+2log_{10}(2) ?
The solution to 2log_{10}(12)-3log_{10}(6)+2log_{10}(2) is log_{10}(8/3)