Relembrando:
[tex]\boxed{\log_k{a^n}=n\log_k{a}}[/tex]
[tex]\boxed{\log_k{\dfrac{a}{b}}=\log_{k}{a}-\log_k{b}}[/tex]
Assim:
[tex]\log_2{\sqrt{\dfrac{4a\sqrt{ab}}{b\sqrt[3]{a^2b}}}}=\log_2{\bigg(\dfrac{4a(a^{\frac{1}{2}}b^{\frac{1}{2}})}{b(a^{\frac{2}{3}}b^{\frac{1}{3}})}\bigg)^{\tfrac{1}{2}}}=[/tex]
[tex]=\dfrac{1}{2}\log_2{\bigg(\dfrac{4a^{\frac{3}{2}}b^{\frac{1}{2}}}{a^{\frac{2}{3}}b^{\frac{4}{3}}}\bigg)}=\dfrac{1}{2}\log_2{\big(4a^{(\frac{3}{2}-\frac{2}{3})}b^{(\frac{1}{2}-\frac{4}{3})}\big)}=[/tex]
[tex]=\dfrac{1}{2}\log_2{\big(4a^{\frac{5}{6}}b^{-\frac{5}{6}}\big)}=\dfrac{1}{2}\bigg(\log_2{4}+\log_2{a^{\frac{5}{6}}}+\log_2{b^{-\frac{5}{6}}\bigg)=[/tex]
[tex]=\dfrac{1}{2}\bigg(2+\dfrac{5}{6}\log_2{a}-\dfrac{5}{6}\log_2{b}\bigg)=\boxed{1+\dfrac{5}{12}\log_2{a}-\dfrac{5}{12}\log_2{b}}[/tex]
Realmente, o gabarito do seu livro parece estar errado.
