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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) \(\ln\left(\sqrt{5}+2\right)+\ln\left(\sqrt{5}-2\right)=ln\left(\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)\right)=\ln\left(\left(\sqrt{5}\right)^2-2^2\right)=ln\left(5-4\right)=\ln1=\ln e^0=1\)
b) \(\log400-\log4=\log\dfrac{400}{4}=\log100=\log10^{10}=10.\log10=10.1=10\)
c) \(\log_48+\log_412+\log_4\dfrac{32}{2}=\log_4\left(8.12.\dfrac{32}{2}\right)=\log_4\left(1024\right)=\log_44^5=5.\log_44=5.1=5\)
a: \(=ln_2\left[\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)\right]=ln1=0\)
b: \(=log\left(\dfrac{400}{4}\right)=log\left(100\right)=10\)
c: \(=log_4\left(8\cdot12\cdot\dfrac{32}{3}\right)=log_4\left(32\cdot32\right)=5\)
\(\dfrac{a^2\cdot\sqrt[3]{a}\cdot\sqrt[5]{a^4}}{\sqrt[4]{a}}=\dfrac{a^2\cdot a^{\dfrac{1}{3}}\cdot a^{\dfrac{4}{5}}}{a^{\dfrac{1}{4}}}=\dfrac{a^{\dfrac{47}{15}}}{a^{\dfrac{1}{4}}}=a^{\dfrac{173}{60}}\)
\(\Rightarrow log_a\left(\dfrac{a^2\cdot\sqrt[3]{a}\cdot\sqrt[5]{a^4}}{\sqrt[4]{a}}\right)=log_a\left(a^{\dfrac{173}{60}}\right)=\dfrac{173}{60}\)
\(a^{2log_a\left(\dfrac{\sqrt{105}}{30}\right)}=a^{log_a\left(\dfrac{7}{60}\right)}=\dfrac{7}{60}\)
Vậy \(B=\dfrac{173}{60}+\dfrac{7}{60}=\dfrac{180}{60}=3\)
\(log_5125=log_55^3=3\)
\(log_6216=log_66^3=3\)
\(log_{10}\dfrac{1}{10000}=log_{10}10^{-4}=-4\)
\(log\sqrt{1000}=log_{10}10^{\dfrac{3}{2}}=\dfrac{3}{2}\)
\(81^{log_35}=3^{3log_35}=3^{log_3125}=125\)
\(125^{log_52}=5^{3log_52}=5^{log_58}=8\)
\(\left(\dfrac{1}{49}\right)^{log_7\dfrac{1}{8}}=7^{-2log_7\dfrac{1}{8}}=7^{log_764}=64\)
\(\left(\dfrac{1}{625}\right)^{log_52}=5^{-4log_52}=5^{log_5\dfrac{1}{16}}=\dfrac{1}{16}\)
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) \(log_3\sqrt[3]{3}=\dfrac{1}{2}\)
b) \(log_{\dfrac{1}{2}}8=-3\)
c) \(\left(\dfrac{1}{25}\right)^{log_54}=\dfrac{1}{16}\)
a: \(=log_3\left(\sqrt{3}\right)^3=log_3\left(3^{\dfrac{1}{2}}\right)^3=log_3\left(3^{\dfrac{3}{2}}\right)=\dfrac{3}{2}\)
b: \(log_{\dfrac{1}{2}}32=log_{\dfrac{1}{2}}\left(\dfrac{1}{2}\right)^{-5}=-5\)
Hàm số a,b là các hàm số logarit
a: \(log_{\sqrt{3}}x\)
Cơ số là \(\sqrt{3}\)
b: \(log_{2^{-2}}x\)
Cơ số là \(2^{-2}=\dfrac{1}{4}\)
a) \(log_69+log_64=log_636=2\)
b) \(log_52-log_550=log_5\left(2:50\right)=-2\)
c) \(log_3\sqrt{5}-\dfrac{1}{2}log_550=-1,0479\)
\(log_3\sqrt{3}=log_33^{\dfrac{1}{2}}=\dfrac{1}{2}\)
\(lne^3=log_ee^3=3\)
\(log_{27}3=log_{3^3}3=\dfrac{1}{3}\)
\(\log_{\sqrt{3}}3=log_{3^{\dfrac{1}{2}}}3=1:\dfrac{1}{2}=2\)
\(\log_{0,125}2=log_{2^{-3}}2=\dfrac{1}{-3}\)
\(\log_{\sqrt[3]{49}}7=\log_{7^{\dfrac{2}{3}}}7=1:\dfrac{2}{3}=\dfrac{3}{2}\)
\(\log_{\dfrac{1}{125}}5=\log_{5^{-3}}5=-\dfrac{1}{3}\)
\(\log_84=log_{2^3}2^2=\dfrac{1}{3}\cdot2=\dfrac{2}{3}\)
\(\log_{25}\left(\dfrac{1}{5}\right)=\log_{5^2}5^{-1}=\dfrac{1}{2}\cdot\left(-1\right)=-\dfrac{1}{2}\)
\(\log_{\dfrac{1}{5}}\sqrt{5}=\log_{5^{-1}}5^{\dfrac{1}{2}}=\dfrac{1}{-1}\cdot\dfrac{1}{2}=-\dfrac{1}{2}\)
\(log_{\dfrac{1}{7}}\sqrt[5]{49}=\log_{7^{-1}}7^{\dfrac{2}{5}}=\dfrac{1}{-1}\cdot\dfrac{2}{5}=-\dfrac{2}{5}\)
\(\log_4\left(\dfrac{1}{\sqrt{2}}\right)=\log_{2^2}\left(\sqrt{2}\right)^{-1}\)
\(=\log_{2^{-2}}\left(\sqrt{2}\right)^{-\dfrac{1}{2}}=\dfrac{1}{-2}\cdot\dfrac{-1}{2}=\dfrac{1}{4}\)
\(\log_{27}3\sqrt{3}=\log_{3^3}3^{\dfrac{3}{2}}=\dfrac{1}{3}\cdot\dfrac{3}{2}=\dfrac{1}{2}\)