What is log_{10}(z)-log_{10}(12x) ?

The solution to log_{10}(z)-log_{10}(12x) is log_{10}(z/(12x))

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simplify log_{10}(z)-log_{10}(12x)

simplify

lo g_{10} (z) - lo g_{10} (12 x)

lo g_{10} (\frac{z}{12 x})

Solution steps

lo g_{10} (z) - lo g_{10} (12 x)

Apply log rule:

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

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

= lo g_{10} (\frac{z}{12 x})

ln (x) - \frac{x}{θ} - 2 ln (θ)

- \frac{7}{12} lo g_{2} (\frac{7}{12}) - \frac{5}{12} lo g_{2} (\frac{5}{12})

ln (x^{5} y^{3})

lo g_{10} (\frac{x}{2 y})

What is log_{10}(z)-log_{10}(12x) ?
The solution to log_{10}(z)-log_{10}(12x) is log_{10}(z/(12x))