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) \(log_69+log_64=log_636=2\)

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

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

Lời giải:

Đặt \(\log_ab=x\Rightarrow \log_ba=\frac{1}{x}\)

a)

\(A=(x+\frac{1}{x}+2)(x-\frac{1}{x}).\frac{1}{x}\)

\(\Leftrightarrow A=(1+\frac{1}{x^2}+2x)(x-\frac{1}{x})=\left(1+\frac{1}{x}\right)^2(x-\frac{1}{x})\)

\(\Leftrightarrow A=(1+\log_ba)^2(\log_ab-\log_ba)\)

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b) Điều kiện: \(x>0\)

Có \(1=\log_{ab}b.\log_b(ab)=\log_{ab}b(\log_ba+\log_bb)=\log_{ab}b(\frac{1}{x}+1)\)

\(\Rightarrow \log_{ab}b=\frac{x}{x+1}\)

Như vậy:

\(B=\sqrt{x+\frac{1}{x}+2}(x-\frac{x}{x+1})\sqrt{x}\)

\(\Leftrightarrow B=\sqrt{x^2+1+2x}(x-\frac{x}{x+1})=|x+1|.\frac{x^2}{x+1}\)

\(=(x+1)\frac{x^2}{x+1}=x^2=\log_a^2b\) (do \(x>0)\)

Chọn A.

Dễ thấy u_n là cấp số nhân với q = 10

Ta có: u₈ = 10⁷u₁; u₁₀ = 10⁹u₁

Do đó PT 

Giải PT ta được logu₁ = -17 ⇔ u₁ = 10^-17 ⇒ u₂₀₁₈ = 10²⁰¹⁷ u₁ = 10²⁰⁰⁰ 

\({a^{\frac{1}{2}}} = b \Leftrightarrow {\log _a}b = \frac{1}{2} \Leftrightarrow 2{\log _a}b = 1\)

Chọn B.

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

\(=log_35^2-log_350+log_36\)

\(=log_3\left(\dfrac{25}{50}\cdot6\right)=log_33=1\)

\(2\log_35-\log_350+\dfrac{1}{2}\log_336=\log_35^2-\log_350+\log_336^{\dfrac{1}{2}}=\log_325-\log_350+\log_36=\log_3\left(\dfrac{25}{50}.6\right)=\log_33=1\)

\(log_3\sqrt{3}=log_33^{\dfrac{1}{2}}=\dfrac{1}{2}\)

\(lne^3=log_ee^3=3\)

\(log_{27}3=log_{3^3}3=\dfrac{1}{3}\)

\(\log_{\sqrt{3}}3=log_{3^{\dfrac{1}{2}}}3=1:\dfrac{1}{2}=2\)

\(\log_{0,125}2=log_{2^{-3}}2=\dfrac{1}{-3}\)

\(\log_{\sqrt[3]{49}}7=\log_{7^{\dfrac{2}{3}}}7=1:\dfrac{2}{3}=\dfrac{3}{2}\)

\(\log_{\dfrac{1}{125}}5=\log_{5^{-3}}5=-\dfrac{1}{3}\)

\(\log_84=log_{2^3}2^2=\dfrac{1}{3}\cdot2=\dfrac{2}{3}\)

\(\log_{25}\left(\dfrac{1}{5}\right)=\log_{5^2}5^{-1}=\dfrac{1}{2}\cdot\left(-1\right)=-\dfrac{1}{2}\)

\(\log_{\dfrac{1}{5}}\sqrt{5}=\log_{5^{-1}}5^{\dfrac{1}{2}}=\dfrac{1}{-1}\cdot\dfrac{1}{2}=-\dfrac{1}{2}\)

\(log_{\dfrac{1}{7}}\sqrt[5]{49}=\log_{7^{-1}}7^{\dfrac{2}{5}}=\dfrac{1}{-1}\cdot\dfrac{2}{5}=-\dfrac{2}{5}\)

\(\log_4\left(\dfrac{1}{\sqrt{2}}\right)=\log_{2^2}\left(\sqrt{2}\right)^{-1}\)

\(=\log_{2^{-2}}\left(\sqrt{2}\right)^{-\dfrac{1}{2}}=\dfrac{1}{-2}\cdot\dfrac{-1}{2}=\dfrac{1}{4}\)

\(\log_{27}3\sqrt{3}=\log_{3^3}3^{\dfrac{3}{2}}=\dfrac{1}{3}\cdot\dfrac{3}{2}=\dfrac{1}{2}\)

a: \(log_22^{-13}=-13\)

b: \(lne^{\sqrt{2}}=\sqrt{2}\)

c: \(log_816-log_82=log_8\left(\dfrac{16}{2}\right)=log_88=1\)

c: \(log_26\cdot log_68=log_28=3\)