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

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

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

a) \(\log_4\sqrt[5]{16}=\log_4\left(4^2\right)^{\dfrac{1}{5}}=\log_44^{\dfrac{2}{5}}=\dfrac{2}{5}\log_44=\dfrac{2}{5}.1=\dfrac{2}{5}\)

b) \(36^{\log_68}=\left(6^2\right)^{\log_68}=6^{2\log_68}=6^{\log_68^2}=8^2=64\)

a: \(log_4\sqrt[5]{16}=log_4\sqrt[5]{4^2}=\dfrac{2}{5}\)

b: \(36^{log_68}=6\cdot^{2\cdot log_68}=8^2=64\)

a: \(=\dfrac{5}{4}\cdot3=\dfrac{15}{4}\)

b: \(=\sqrt[5]{\dfrac{98}{64}\cdot343}=\sqrt[5]{\left(\dfrac{7}{2}\right)^5}=\dfrac{7}{2}\)

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: \(=log_4\left(8\cdot12\cdot\dfrac{32}{3}\right)=log_4\left(32\cdot32\right)=5\)

a, Điều kiện: x > 0

\(log_3\left(x\right)< 2\\ \Rightarrow0< x< 9\)

b, Điều kiện: x > 5

\(log_{\dfrac{1}{4}}\left(x-5\right)\ge-2\\ \Rightarrow x-5\le16\\ \Leftrightarrow5< x\le21\)

a) \(\log_381=\log_33^4=4\log_33=4.1=4\)

b) \(\log_{10}\dfrac{1}{100}=\log_{10}10^{-2}=-2\log_{10}10=-2.1=-2\)

a: \(log_381=4\)

b: \(log_{10}\left(\dfrac{1}{100}\right)=-2\)

\(log_719\simeq1,51;log_{11}26\simeq1,36\)

Bảng biến thiên:

Đồ thị: 

2;-2 là căn bậc 6 của 64 vì \(2^6=64;\left(-2\right)^6=64\)