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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

2 lo g_{5} (2 x - 1) = 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 log rule:

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

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

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

Apply log rule:

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

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

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

Apply the fraction rule:

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

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

ln (x) + ln (18)

ln (a + b) + ln (a - b) - 4 ln (c)

lo g_{10} (\frac{5 y}{x ^{2}})

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

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