What is log_{10}(3x+1)-log_{10}(3x^2-5x-2) ?

The solution to log_{10}(3x+1)-log_{10}(3x^2-5x-2) is -log_{10}(x-2)

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simplify log_{10}(3x+1)-log_{10}(3x^2-5x-2)

simplify

lo g_{10} (3 x + 1) - lo g_{10} (3 x^{2} - 5 x - 2)

- lo g_{10} (x - 2)

Solution steps

lo g_{10} (3 x + 1) - lo g_{10} (3 x^{2} - 5 x - 2)

Apply log rule:

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

lo g_{10} (3 x + 1) - lo g_{10} (3 x^{2} - 5 x - 2) = lo g_{10} (\frac{3 x + 1}{3 x ^{2} - 5 x - 2})

= lo g_{10} (\frac{3 x + 1}{3 x ^{2} - 5 x - 2})

Simplify

\frac{3 x + 1}{3 x ^{2} - 5 x - 2} : \frac{1}{x - 2}

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

Simplify

\frac{1}{x - 2} : (x - 2)^{- 1}

= lo g_{10} ((x - 2)^{- 1})

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x), assuming x > 0

lo g_{10} ((x - 2)^{- 1}) = - 1 \cdot lo g_{10} (x - 2)

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

Apply rule:

1 \cdot a = a

= - lo g_{10} (x - 2)

ln (2 e^{1})

lo g_{6} (\frac{w ^{5}}{v})

\frac{1}{2} \cdot ln (3)

lo g_{16} (\frac{9 r}{s})

lo g_{10} (a - b) - lo g_{10} (3)

What is log_{10}(3x+1)-log_{10}(3x^2-5x-2) ?
The solution to log_{10}(3x+1)-log_{10}(3x^2-5x-2) is -log_{10}(x-2)