What is log_{10}(x-1)-log_{10}(x+2) ?

The solution to log_{10}(x-1)-log_{10}(x+2) is log_{10}((x-1)/(x+2))

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simplify log_{10}(x-1)-log_{10}(x+2)

simplify

lo g_{10} (x - 1) - lo g_{10} (x + 2)

lo g_{10} (\frac{x - 1}{x + 2})

Solution steps

lo g_{10} (x - 1) - lo g_{10} (x + 2)

Apply log rule:

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

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

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

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What is log_{10}(x-1)-log_{10}(x+2) ?
The solution to log_{10}(x-1)-log_{10}(x+2) is log_{10}((x-1)/(x+2))