Помогите решить 2 и 4 задание по логарифмам

Решите задачу:

$1)\; \; ( \frac{3ab}{5cd^{-1}} )^4\cdot ( \frac{ac^{-4}}{b^2d^3} )^{-2}\cdot ( \frac{a^{-2}b^2}{cd^{-3}} )^4= \frac{3^4a^4b^4d^4}{5^4c^4} \cdot \frac{c^8b^4d^6}{a^2} \cdot \frac{b^8d^{12}}{a^8c^4} =\\\\= \frac{81a^4b^{16}c^8d^{22}}{625a^{10}c^8} = \frac{81b^{16}d^{22}}{625a^6}$

$2)\; \; log_{ab} \frac{\sqrt{b}}{a}+log_{\sqrt{ab}}b+log_{a}{\sqrt[3]{b}}= \frac{log_{b}\frac{\sqrt{b}}{a}}{log_{b}(ab)} +\frac{1}{log_{b}\sqrt{ab}}+ \frac{log_{b}\sqrt[3]{b}}{log_{b}{a}} =\\\\= \frac{log_b\sqrt{b}-log_{b}a}{log_{b}a+log_{b}b}+\frac{1}{log_{b}\sqrt{a}+log_{b}\sqrt{b}} +\frac{\frac{1}{3}}{2}= \frac{\frac{1}{2}-2}{2+1} + \frac{1}{\frac{1}{2}log_{b}a+\frac{1}{2}} +\frac{1}{6}=\\\\=\frac{-\frac{3}{2}}{3}+\frac{1}{\frac{1}{2}\cdot 2+\frac{1}{2}}+\frac{1}{6}=\\\\=-\frac{1}{2}+\frac{2}{3}+\frac{1}{6}=0$

$log_{b}a=2$