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

The solution to expand log_{4}(x^2-1)-5log_{4}(x+1) is log_{4}((x-1)/(x^4+4x^3+6x^2+4x+1))

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expand log_{4}(x^2-1)-5log_{4}(x+1)

expand

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

lo g_{4} (\frac{x - 1}{x ^{4} + 4 x ^{3} + 6 x ^{2} + 4 x + 1})

Solution steps

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

Apply log rule:

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

5 lo g_{4} (x + 1) = lo g_{4} ((x + 1)^{5})

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

Apply log rule:

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

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

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

Expand

\frac{x ^{2} - 1}{( x + 1 ) ^{5}} : \frac{x - 1}{x ^{4} + 4 x ^{3} + 6 x ^{2} + 4 x + 1}

= lo g_{4} (\frac{x - 1}{x ^{4} + 4 x ^{3} + 6 x ^{2} + 4 x + 1})

lo g_{3} (\frac{100}{27 x})

ln (\frac{4 e ^{6}}{7 y})

lo g_{7} (x) - lo g_{7} (x^{2})

ln (\frac{x}{b})

What is expand log_{4}(x^2-1)-5log_{4}(x+1) ?
The solution to expand log_{4}(x^2-1)-5log_{4}(x+1) is log_{4}((x-1)/(x^4+4x^3+6x^2+4x+1))