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) \(log_69+log_64=log_636=2\)

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) \(log_216=4\)

b) \(log_3\dfrac{1}{27}=-3\)

c) \(log1000=3\)

d) \(9^{log_312}=144\)

\(a,A=log_23\cdot log_34\cdot log_45\cdot log_56\cdot log_67\cdot log_78\\ =log_28\\ =log_22^3\\ =3\\ b,B=log_22\cdot log_24...log_22^n\\ =log_22\cdot log_22^2...log_22^n\\ =1\cdot2\cdot...\cdot n\\ =n!\)

a) \(\sqrt[4]{\dfrac{1}{16}}=\dfrac{1}{2}\)

b) \(\left(\sqrt[6]{8}\right)^2=\sqrt[\dfrac{6}{2}]{8}=\sqrt[3]{8}=2\)

c) \(\sqrt[4]{3}\cdot\sqrt[4]{27}=\sqrt[4]{3\cdot27}=\sqrt[4]{81}=3\)

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

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

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

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

a) Với \(x = 1\) thì \(y = {\log _2}1 = 0\)

Với \(x = 2\) thì \(y = {\log _2}2 = 1\)

Với \(x = 4\) thì \(y = {\log _2}4 = 2\)

b) Biểu thức \(y = {\log _2}x\) có nghĩa khi x > 0.

a)

$16^{\alpha }+16^{-\alpha } = (4^2)^{\alpha }+(4^2)^{-\alpha } = 4^{2\alpha }+4^{-2\alpha }$

$4^{2\alpha }+4^{-2\alpha } = 4^{2\log_4{\frac{1}{5}}}+4^{-2\log_4{\frac{1}{5}}} = \left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^{-2} = \frac{1}{25}+25 = \frac{26}{25}$

b)

$\left(2^{\alpha }+2^{-\alpha }\right)^2 = \left(\sqrt{4}\right)^{\alpha }+\left(\sqrt{4}\right)^{-\alpha } = 4^{\frac{\alpha}{2}}+4^{-\frac{\alpha}{2}}$

$4^{\frac{\alpha}{2}}+4^{-\frac{\alpha}{2}} = 4^{\frac{\log_4{\frac{1}{5}}}{2}}+4^{-\frac{\log_4{\frac{1}{5}}}{2}} = \left(\frac{1}{5}\right)^{\frac{1}{2}}+\left(\frac{1}{5}\right)^{-\frac{1}{2}} = \sqrt{\frac{1}{5}}+\frac{1}{\sqrt{5}} = \frac{2}{\sqrt{5}}$