What is log_{10}(2x)-log_{10}(8x^3) ?

The solution to log_{10}(2x)-log_{10}(8x^3) is -log_{10}(4x^2)

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simplify log_{10}(2x)-log_{10}(8x^3)

simplify

lo g_{10} (2 x) - lo g_{10} (8 x^{3})

- lo g_{10} (4 x^{2})

Solution steps

lo g_{10} (2 x) - lo g_{10} (8 x^{3})

Apply log rule:

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

lo g_{10} (2 x) - lo g_{10} (8 x^{3}) = lo g_{10} (\frac{2 x}{8 x ^{3}})

= lo g_{10} (\frac{2 x}{8 x ^{3}})

Cancel

\frac{2 x}{8 x ^{3}} : \frac{1}{4 x ^{2}}

= lo g_{10} (\frac{1}{4 x ^{2}})

Simplify

\frac{1}{4 x ^{2}} : (4 x^{2})^{- 1}

= lo g_{10} ((4 x^{2})^{- 1})

Apply log rule

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

lo g_{10} ((4 x^{2})^{- 1}) = - 1 \cdot lo g_{10} (4 x^{2})

= - 1 \cdot lo g_{10} (4 x^{2})

Apply rule:

1 \cdot a = a

= - lo g_{10} (4 x^{2})

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What is log_{10}(2x)-log_{10}(8x^3) ?
The solution to log_{10}(2x)-log_{10}(8x^3) is -log_{10}(4x^2)