a) \(log_29\cdot log_34=4\)

b) \(log_{25}\cdot\dfrac{1}{\sqrt{5}}=-\dfrac{1}{4}\)

c) \(log_23\cdot log_9\sqrt{5}\cdot log_54=\dfrac{1}{2}\)

a) \(log_3\sqrt[3]{3}=\dfrac{1}{2}\)

b) \(log_{\dfrac{1}{2}}8=-3\)

c) \(\left(\dfrac{1}{25}\right)^{log_54}=\dfrac{1}{16}\)

a, Hàm số \(y=log_{\dfrac{1}{2}}x\) có cơ số \(\dfrac{1}{2}< 1\) nên hàm số nghịch biến trên \(\left(0;+\infty\right)\)

Mà \(4,8< 5,2\Rightarrow log_{\dfrac{1}{2}}4,8>log_{\dfrac{1}{2}}5,2\)

b, Ta có: \(log_{\sqrt{5}}2=2log_52=log_54\)

Hàm số \(y=log_5x\) có cơ số 5 > 1 nên hàm số đồng biến trên \(\left(0;+\infty\right)\)

Do \(4>2\sqrt{2}\Rightarrow log_54>log_52\sqrt{2}\Rightarrow log_{\sqrt{5}}2>log_52\sqrt{2}\)

c, Ta có: \(-log_{\dfrac{1}{4}}2=-\dfrac{1}{2}log_{\dfrac{1}{2}}2=log_{\dfrac{1}{2}}\dfrac{1}{\sqrt{2}}\)

Hàm số \(y=log_{\dfrac{1}{2}}x\) có cơ số \(\dfrac{1}{2}< 1\) nên nghịch biến trên \(\left(0;+\infty\right)\)

Do \(\dfrac{1}{\sqrt{2}}>0,4\Rightarrow log_{\dfrac{1}{2}}\dfrac{1}{\sqrt{2}}< log_{\dfrac{1}{2}}0,4\Rightarrow-log_{\dfrac{1}{4}}2< log_{\dfrac{1}{2}}0,4\)

a, Điều kiện: \(2^x\ne3\Rightarrow x\ne log_23\)

Vậy D = R \ \(log_23\)

b, Điều kiện: \(25-5^x\ge0\Rightarrow5^x\le5^2\Rightarrow x\le2\)

Vậy D = \((-\infty;2]\)

c, Điều kiện: \(\left\{{}\begin{matrix}x>0\\lnx\ne1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>0\\x\ne e\end{matrix}\right.\)

Vậy D = \(\left(0;+\infty\right)\backslash\left\{e\right\}\)

d, Điều kiện: \(\left\{{}\begin{matrix}x>0\\1-log_3x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>0\\log_3x\le1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>0\\x\le3\end{matrix}\right.\Rightarrow0< x\le3\)

Vậy D = \((0;3]\)

\(\dfrac{a^2\cdot\sqrt[3]{a}\cdot\sqrt[5]{a^4}}{\sqrt[4]{a}}=\dfrac{a^2\cdot a^{\dfrac{1}{3}}\cdot a^{\dfrac{4}{5}}}{a^{\dfrac{1}{4}}}=\dfrac{a^{\dfrac{47}{15}}}{a^{\dfrac{1}{4}}}=a^{\dfrac{173}{60}}\)

\(\Rightarrow log_a\left(\dfrac{a^2\cdot\sqrt[3]{a}\cdot\sqrt[5]{a^4}}{\sqrt[4]{a}}\right)=log_a\left(a^{\dfrac{173}{60}}\right)=\dfrac{173}{60}\)

\(a^{2log_a\left(\dfrac{\sqrt{105}}{30}\right)}=a^{log_a\left(\dfrac{7}{60}\right)}=\dfrac{7}{60}\)

Vậy \(B=\dfrac{173}{60}+\dfrac{7}{60}=\dfrac{180}{60}=3\)

a) \(log_69+log_64=log_636=2\)

b) \(log_52-log_550=log_5\left(2:50\right)=-2\)

c) \(log_3\sqrt{5}-\dfrac{1}{2}log_550=-1,0479\)

a,Ta có: \(a^6=\left(a^{\dfrac{3}{5}}\right)^{10}=b^{10}\\ a^3b=\left(a^{\dfrac{3}{5}}\right)^5\cdot b=b^5\cdot b=b^6\\ \dfrac{a^9}{b^9}=\dfrac{\left(a^{\dfrac{3}{5}}\right)^{15}}{b^9}=\dfrac{b^{15}}{b^9}=b^6\)

b, \(log_ab=log_aa^{\dfrac{3}{5}}=\dfrac{3}{5}\\ log_a\left(a^2b^5\right)=log_a\left(a^2\cdot a^3\right)=log_a\left(a^5\right)=5\\ log_{\sqrt[5]{a}}\left(\dfrac{a}{b}\right)=5log_a\left(\dfrac{a}{a^{\dfrac{3}{5}}}\right)=5log_a\left(a^{\dfrac{2}{5}}\right)=2\)