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

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

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

simplify

\frac{( lo g _{10} ( 3 x ) )}{( x ^{2} + 1 )}

\frac{lo g _{10} ( 3 ) + lo g _{10} ( x )}{x ^{2} + 1}

Solution steps

\frac{lo g _{10} ( 3 x )}{x ^{2} + 1}

Apply log rule:

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

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

= \frac{lo g _{10} ( 3 ) + lo g _{10} ( x )}{x ^{2} + 1}

(a^{\frac{3}{5}}) (a^{\frac{12}{5}})

(2 a + 1) (a - 1)

2 m + 30 - 41.3 m - n + 5

(2 x + 9) (2 x + 4)

(m - 10)^{3}

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