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.

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

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

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

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

\(P=loga^3+logb^2=log\left(a^3b^2\right)=log\left(100\right)=10\)

Chọn B

\(=\left(log_{a^{-1}}a^2\right)^2+\dfrac{1}{2}.\dfrac{1}{2}log_aa\)

\(=\left(-1.2.log_aa\right)^2+\dfrac{1}{4}=4+\dfrac{1}{4}=\dfrac{17}{4}\)

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

Đặt  x = a - 1 6 ;  y = b - 1 6 . Khi đó

K = x + y x 2 - x y + y 2 x 3 - y 3 = x 3 + y 3 x 3 - y 3 = x 6 - y 6 = a - 1 - b = 1 - a b a

Đáp án D

Đáp án là C

Chọn C.