What is 1+log_{x}(x+1)-log_{x}(x+4) ?

The solution to 1+log_{x}(x+1)-log_{x}(x+4) is 1+log_{x}((x+1)/(x+4))

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simplify 1+log_{x}(x+1)-log_{x}(x+4)

simplify

1 + lo g_{x} (x + 1) - lo g_{x} (x + 4)

1 + lo g_{x} (\frac{x + 1}{x + 4})

Solution steps

1 + lo g_{x} (x + 1) - lo g_{x} (x + 4)

Apply log rule:

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

lo g_{x} (x + 1) - lo g_{x} (x + 4) = lo g_{x} (\frac{x + 1}{x + 4})

= 1 + lo g_{x} (\frac{x + 1}{x + 4})

lo g_{3} (5 x y)

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

ln (\frac{p N ^{\frac{1}{r}}}{K ^{\frac{1}{r}} ( - p + 1 )})

\frac{1}{2} (lo g_{6} (a) + lo g_{6} (b))

\frac{1}{2} lo g_{9} (18) - lo g_{9} (2)

What is 1+log_{x}(x+1)-log_{x}(x+4) ?
The solution to 1+log_{x}(x+1)-log_{x}(x+4) is 1+log_{x}((x+1)/(x+4))