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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply rule

1^{a} = 1

1^{3} = 1

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

Apply rule

1^{a} = 1

1^{2} = 1

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

ln (\frac{7163}{7})

ln (\frac{12 \cdot 1 0 ^{6}}{1 \cdot 1 0 ^{8}})

1 + ln (\frac{1}{2})

ln (3.8 2.2 x + 5.4)

5 ln (\frac{5}{3})

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