What is 2log_{a}(z+1)+log_{a}(3z+6) ?

The solution to 2log_{a}(z+1)+log_{a}(3z+6) is log_{a}((z+1)^2(3z+6))

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simplify 2log_{a}(z+1)+log_{a}(3z+6)

simplify

2 lo g_{a} (z + 1) + lo g_{a} (3 z + 6)

lo g_{a} ((z + 1)^{2} (3 z + 6))

Solution steps

2 lo g_{a} (z + 1) + lo g_{a} (3 z + 6)

Apply log rule:

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

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

= lo g_{a} ((z + 1)^{2}) + lo g_{a} (3 z + 6)

Apply log rule:

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

lo g_{a} ((z + 1)^{2}) + lo g_{a} (3 z + 6) = lo g_{a} ((z + 1)^{2} (3 z + 6))

= lo g_{a} ((z + 1)^{2} (3 z + 6))

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ln (\frac{e ^{i 0}}{4 + 3 e ^{i 0}})

\frac{7}{4} ln ∣ x ∣ - \frac{3}{4} ln ∣ x + 4 ∣

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What is 2log_{a}(z+1)+log_{a}(3z+6) ?
The solution to 2log_{a}(z+1)+log_{a}(3z+6) is log_{a}((z+1)^2(3z+6))