What is 3log_{10}(3)+log_{10}(x)-log_{10}(3x) ?

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

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simplify 3log_{10}(3)+log_{10}(x)-log_{10}(3x)

simplify

3 lo g_{10} (3) + lo g_{10} (x) - lo g_{10} (3 x)

2 lo g_{10} (3)

Decimal Notation

0.95424 \dots

Solution steps

3 lo g_{10} (3) + lo g_{10} (x) - lo g_{10} (3 x)

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Simplify

3^{3} x : 27 x

= lo g_{10} (27 x) - lo g_{10} (3 x)

Apply log rule:

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

lo g_{10} (27 x) - lo g_{10} (3 x) = lo g_{10} (\frac{27 x}{3 x})

= lo g_{10} (\frac{27 x}{3 x})

Cancel

\frac{27 x}{3 x} : 9

= lo g_{10} (9)

Factor the number:

9 = 3^{2}

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

Apply log rule:

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

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

= 2 lo g_{10} (3)

lo g_{10} (7) - lo g_{10} (4)

ln (9 x (2 x + 5))

ln (\frac{11}{4})

lo g_{b} (x y)

lo g_{6} (x) - 2 lo g_{6} (y) + lo g_{6} (z)

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