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

The solution to (2)log_{10}(3x^2)-log_{10}(2x^2-1) is log_{10}((9x^4)/(2x^2-1))

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

simplify

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

lo g_{10} (\frac{9 x ^{4}}{2 x ^{2} - 1})

Solution steps

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

Apply rule:

(a) = a

(2) = 2

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

Apply log rule:

b \cdot lo g_{a} (x) = lo g_{a} (x^{b})

2 lo g_{10} (3 x^{2}) = lo g_{10} ((3 x^{2})^{2})

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

Apply log rule:

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

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

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

Simplify

(3 x^{2})^{2} : 9 x^{4}

= lo g_{10} (\frac{9 x ^{4}}{2 x ^{2} - 1})

3 lo g_{5} (a) - lo g_{5} (b)

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

lo g_{8} (\frac{64}{x + 1})

9 ln (a) - 6 ln (b)

What is (2)log_{10}(3x^2)-log_{10}(2x^2-1) ?
The solution to (2)log_{10}(3x^2)-log_{10}(2x^2-1) is log_{10}((9x^4)/(2x^2-1))