Question

Помогите решить 2 и 4 задание по логарифмам

Answer 1

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

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

$log_{b}a=2$