What is log_{b}(x+1)+log_{b}(3)-log_{b}(x+2) ?

The solution to log_{b}(x+1)+log_{b}(3)-log_{b}(x+2) is log_{b}((3(x+1))/(x+2))

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simplify log_{b}(x+1)+log_{b}(3)-log_{b}(x+2)

simplify

lo g_{b} (x + 1) + lo g_{b} (3) - lo g_{b} (x + 2)

lo g_{b} (\frac{3 ( x + 1 )}{x + 2})

Solution steps

lo g_{b} (x + 1) + lo g_{b} (3) - lo g_{b} (x + 2)

Apply log rule:

lo g_{a} (x) + lo g_{a} (y) = lo g_{a} (x y)

lo g_{b} (x + 1) + lo g_{b} (3) = lo g_{b} ((x + 1) \cdot 3)

= lo g_{b} ((x + 1) \cdot 3) - lo g_{b} (x + 2)

Apply log rule:

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

lo g_{b} ((x + 1) \cdot 3) - lo g_{b} (x + 2) = lo g_{b} (\frac{( x + 1 ) \cdot 3}{x + 2})

= lo g_{b} (\frac{( x + 1 ) \cdot 3}{x + 2})

= lo g_{b} (\frac{3 ( x + 1 )}{x + 2})

lo g_{4} (\frac{16}{x + 9})

ln (\frac{x ^{2} - 1}{x ^{3}})

4 lo g_{2} (u) + 7 lo g_{2} (v)

\frac{1}{2} lo g_{c} (x) + 2 lo g_{c} (y) - 3 lo g_{c} (x)

- lo g_{10} (1.9 \cdot 1 0^{- 6})

What is log_{b}(x+1)+log_{b}(3)-log_{b}(x+2) ?
The solution to log_{b}(x+1)+log_{b}(3)-log_{b}(x+2) is log_{b}((3(x+1))/(x+2))