What is log_{a}((a^3+2a^2+a)/((a+a)^2)) ?

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

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simplify log_{a}((a^3+2a^2+a)/((a+a)^2))

simplify

lo g_{a} (\frac{a ^{3} + 2 a ^{2} + a}{( a + a ) ^{2}})

2 lo g_{a} (a + 1) - 2 lo g_{a} (2) - 1

Solution steps

lo g_{a} (\frac{a ^{3} + 2 a ^{2} + a}{( a + a ) ^{2}})

Simplify

\frac{a ^{3} + 2 a ^{2} + a}{( a + a ) ^{2}} : \frac{( a + 1 ) ^{2}}{4 a}

= lo g_{a} (\frac{( a + 1 ) ^{2}}{4 a})

Apply log rule:

lo g_{a} (\frac{x}{y}) = lo g_{a} (x) - lo g_{a} (y)

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

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

lo g_{a} ((a + 1)^{2}) = 2 lo g_{a} (a + 1)

lo g_{a} (4 a) = 2 lo g_{a} (2) + 1

= 2 lo g_{a} (a + 1) - (2 lo g_{a} (2) + 1)

Apply the distributive law:

- (a + b) = - a - b

- (2 lo g_{a} (2) + 1) = - 2 lo g_{a} (2) - 1

= 2 lo g_{a} (a + 1) - 2 lo g_{a} (2) - 1

lo g_{10} (3) (9)

\frac{e ^{- 2} \cdot 2 ^{3}}{3 !}

12 - \frac{2}{5} 75

\frac{4}{7} + \frac{1}{2}

\frac{6 ^{7}}{6 ^{3}}

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