What is log_{10}(2)+log_{10}(6)-log_{10}(4) ?

The solution to log_{10}(2)+log_{10}(6)-log_{10}(4) is log_{10}(3)

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simplify log_{10}(2)+log_{10}(6)-log_{10}(4)

simplify

lo g_{10} (2) + lo g_{10} (6) - lo g_{10} (4)

lo g_{10} (3)

Decimal Notation

0.47712 \dots

Solution steps

lo g_{10} (2) + lo g_{10} (6) - lo g_{10} (4)

Apply log rule:

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

lo g_{10} (2) + lo g_{10} (6) = lo g_{10} (2 \cdot 6)

= lo g_{10} (2 \cdot 6) - lo g_{10} (4)

Multiply the numbers:

2 \cdot 6 = 12

= lo g_{10} (12) - lo g_{10} (4)

Apply log rule:

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

lo g_{10} (12) - lo g_{10} (4) = lo g_{10} (\frac{12}{4})

= lo g_{10} (\frac{12}{4})

Divide the numbers:

\frac{12}{4} = 3

= lo g_{10} (3)

lo g_{2} (x^{3} y^{4})

4 lo g_{b} (x) + 9 lo g_{b} (z)

ln (\frac{7000}{ln ( 4 )})

ln (\frac{4}{3 x ^{3} y})

lo g_{2} (\frac{7 r ^{7}}{z ^{5}})

What is log_{10}(2)+log_{10}(6)-log_{10}(4) ?
The solution to log_{10}(2)+log_{10}(6)-log_{10}(4) is log_{10}(3)