What is 5log_{10}(x)-4log_{10}(x^2+1)+3log_{10}(x-1) ?

The solution to 5log_{10}(x)-4log_{10}(x^2+1)+3log_{10}(x-1) is log_{10}((x^5(x-1)^3)/((x^2+1)^4))

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simplify 5log_{10}(x)-4log_{10}(x^2+1)+3log_{10}(x-1)

simplify

5 lo g_{10} (x) - 4 lo g_{10} (x^{2} + 1) + 3 lo g_{10} (x - 1)

lo g_{10} (\frac{x ^{5} ( x - 1 ) ^{3}}{( x ^{2} + 1 ) ^{4}})

Solution steps

5 lo g_{10} (x) - 4 lo g_{10} (x^{2} + 1) + 3 lo g_{10} (x - 1)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

5 lo g_{10} (x) = lo g_{10} (x^{5})

= lo g_{10} (x^{5}) - 4 lo g_{10} (x^{2} + 1) + 3 lo g_{10} (x - 1)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

4 lo g_{10} (x^{2} + 1) = lo g_{10} ((x^{2} + 1)^{4})

= lo g_{10} (x^{5}) - lo g_{10} ((x^{2} + 1)^{4}) + 3 lo g_{10} (x - 1)

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

lo g_{10} (x^{5}) - lo g_{10} ((x^{2} + 1)^{4}) = lo g_{10} (\frac{x ^{5}}{( x ^{2} + 1 ) ^{4}})

= lo g_{10} (\frac{x ^{5}}{( x ^{2} + 1 ) ^{4}}) + 3 lo g_{10} (x - 1)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

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

= lo g_{10} (\frac{x ^{5}}{( x ^{2} + 1 ) ^{4}}) + lo g_{10} ((x - 1)^{3})

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

lo g_{10} (\frac{x ^{5}}{( x ^{2} + 1 ) ^{4}}) + lo g_{10} ((x - 1)^{3}) = lo g_{10} (\frac{x ^{5}}{( x ^{2} + 1 ) ^{4}} (x - 1)^{3})

= lo g_{10} (\frac{x ^{5}}{( x ^{2} + 1 ) ^{4}} (x - 1)^{3})

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= lo g_{10} (\frac{x ^{5} ( x - 1 ) ^{3}}{( x ^{2} + 1 ) ^{4}})

lo g_{4} (x - 3) + lo g_{4} (x + 3)

lo g_{10} (\frac{x}{1000})

lo g_{10} (\frac{x ^{3}}{y ^{4}})

2 lo g_{10} (b x) + 3 lo g_{10} (b) y

What is 5log_{10}(x)-4log_{10}(x^2+1)+3log_{10}(x-1) ?
The solution to 5log_{10}(x)-4log_{10}(x^2+1)+3log_{10}(x-1) is log_{10}((x^5(x-1)^3)/((x^2+1)^4))