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

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

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

simplify

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

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

Solution steps

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

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

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

lo g_{10} (\frac{x ^{5}}{z y ^{4}})

\frac{3}{2} ln (\frac{3}{2})

5 \cdot ln (\frac{( 300000 + 320 )}{320})

ln (\frac{1}{2}) + \frac{1}{2}

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