What is log_{10}(0.1)-log_{10}(0.2) ?

The solution to log_{10}(0.1)-log_{10}(0.2) is -log_{10}(2)

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simplify log_{10}(0.1)-log_{10}(0.2)

simplify

lo g_{10} (0.1) - lo g_{10} (0.2)

- lo g_{10} (2)

Decimal Notation

- 0.30102 \dots

Solution steps

lo g_{10} (0.1) - lo g_{10} (0.2)

Apply log rule:

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

lo g_{10} (0.1) - lo g_{10} (0.2) = lo g_{10} (\frac{0.1}{0.2})

= lo g_{10} (\frac{0.1}{0.2})

\frac{0.1}{0.2} = \frac{1}{2}

= lo g_{10} (\frac{1}{2})

\frac{1}{2} = 2^{- 1}

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

Apply log rule:

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

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

= - 1 \cdot lo g_{10} (2)

Apply rule:

1 \cdot a = a

= - lo g_{10} (2)

- lo g_{10} (4.7 \cdot 1 0^{- 6})

- lo g_{10} (7.45 \cdot 1 0^{- 6})

ln (- 1 - i) - ln (i)

\frac{ln ( 11000 ) - ln ( 7000 )}{3}

ln (\frac{58387}{88}) \cdot (- \frac{80}{3})

What is log_{10}(0.1)-log_{10}(0.2) ?
The solution to log_{10}(0.1)-log_{10}(0.2) is -log_{10}(2)