What is log_{10}(2)-log_{10}(14) ?

The solution to log_{10}(2)-log_{10}(14) is -log_{10}(7)

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simplify log_{10}(2)-log_{10}(14)

simplify

lo g_{10} (2) - lo g_{10} (14)

- lo g_{10} (7)

Decimal Notation

- 0.84509 \dots

Solution steps

lo g_{10} (2) - lo g_{10} (14)

Apply log rule:

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

lo g_{10} (2) - lo g_{10} (14) = lo g_{10} (\frac{2}{14})

= lo g_{10} (\frac{2}{14})

Cancel the numbers:

\frac{2}{14} = \frac{1}{7}

= lo g_{10} (\frac{1}{7})

\frac{1}{7} = 7^{- 1}

= lo g_{10} (7^{- 1})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x), x > 0

lo g_{10} (7^{- 1}) = - 1 \cdot lo g_{10} (7)

= - 1 \cdot lo g_{10} (7)

Apply rule:

1 \cdot a = a

= - lo g_{10} (7)

lo g_{2} (x^{2} - 3 x + 2) - lo g_{2} (x - 1)

2 lo g_{10} (5) - lo g_{10} (15) + lo g_{10} (9)

lo g_{3} (\frac{3}{y})

lo g_{x} (x y)

lo g_{7} (\frac{2 ^{3}}{5 ^{2}})

What is log_{10}(2)-log_{10}(14) ?
The solution to log_{10}(2)-log_{10}(14) is -log_{10}(7)