What is log_{10}(4)+log_{10}(3)-log_{10}(2) ?

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

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simplify log_{10}(4)+log_{10}(3)-log_{10}(2)

simplify

lo g_{10} (4) + lo g_{10} (3) - lo g_{10} (2)

lo g_{10} (6)

Decimal Notation

0.77815 \dots

Solution steps

lo g_{10} (4) + lo g_{10} (3) - lo g_{10} (2)

Apply log rule:

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

lo g_{10} (4) + lo g_{10} (3) = lo g_{10} (4 \cdot 3)

= lo g_{10} (4 \cdot 3) - lo g_{10} (2)

Multiply the numbers:

4 \cdot 3 = 12

= lo g_{10} (12) - lo g_{10} (2)

Apply log rule:

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

lo g_{10} (12) - lo g_{10} (2) = lo g_{10} (\frac{12}{2})

= lo g_{10} (\frac{12}{2})

Divide the numbers:

\frac{12}{2} = 6

= lo g_{10} (6)

ln (x^{6} y^{3})

- lo g_{10} (2 \cdot 1 0^{- 3})

ln (x) + ln (x - 10)

- lo g_{10} (2 \cdot 1 0^{- 5})

lo g_{10} (x^{3} y^{2})

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