\(\log_2 \displaystyle\frac{1}{4}=\log_2 2^{-2}=-2\).

Ta có:

\(\log_{25} \displaystyle\frac{1}{\sqrt{5}}=\log_{5^2} (5^{\frac{-1}{2}})=\displaystyle\frac{1}{2}\cdot \displaystyle\frac{-1}{2}\log_5 5=-\displaystyle\frac{1}{4}\cdot 1=-\displaystyle\frac{1}{4}\).

\(9^{\log_3 5} =\left(3^2\right)^{\log_3 5}=3^{2\log_3 5}=\left(3^{\log_3 5}\right)^2=5^2=25\).

\(\log_3 \sqrt[3]{3}=\log_3 \left(3\right)^{\frac{1}{3}}=\displaystyle\frac{1}{3}\log_3 3=\displaystyle\frac{1}{3}\).

\(\log_{\frac{1}{2}} 8=\log_{2^{-1}} \left(2\right)^3=-3\log_2 2=-3\).

Ta có: \(\log _4 8=\displaystyle\frac{\log _2 8}{\log _2 4}=\displaystyle\frac{\log _2 2^3}{\log _2 2^2}=\displaystyle\frac{3}{2}\).

\(\log _2 2^{-13}=-13\log_22=-13\).

\(\ln \mathrm{e}^{\sqrt{2}}=\sqrt{2}\ln\mathrm{e}=\sqrt{2}\).

\(\log _2 \displaystyle\frac{1}{8}=\log _2 2^{-3}=-3\).

\(\log _{\sqrt{3}} 9=\log _{\sqrt{3}}(\sqrt{3})^4=4\).

\(8^{\log _{2} 5}=\left(2^3\right)^{\log_25}=\left(2^{\log_25}\right)^3=5^3=125\).

\(\left(\displaystyle\frac{1}{10}\right)^{\log 81}=\left(10^{-1}\right)^{\log 81}=\left(10^{\log 81}\right)^{-1}=81^{-1}=\displaystyle\frac{1}{81}\).

\(5^{\log _{25} 16}=5^{\log _{5^2} 2^4}=5^{2\log _{5} 2}=\left(5^{\log_52}\right)^2=2^2=4\).

\(\log _{12} 12^{3}=3\).

\(\log _{0{,}5} 0,25=\log _{0,5} 0{,}5^2=2\).

\(\log _{9} 3=\log _{3^{2}} 3=\displaystyle\frac{1}{2} \log _{3} 3=\displaystyle\frac{1}{2}\).

\(5^{\log _{125} 64}=5^{\log _{5^3} 2^6}=5^{\frac{6}{3}\log _{5} 2}=5^{2\log _{5} 2}=\left({5^{\log _{5} 2}}\right)^2=2^2=4\).

\(\log _{3} 9^{2}=2 \log _{3} 9=2 \log _{3} 3^{2}=2.2 . \log _{3} 3=4\).

\(\log_{2} 16=\log_2 2^4=4\cdot \log_2 2=4\cdot 1=4\).

\(\log_3 27=\log_3 3^3=3 \cdot \log_3 3=3\cdot 1=3\).

\(\log 1000=\log 10^3=3\cdot \log 10=3\cdot 1=3\).

\(9^{\log_3 12} =\left(3^2\right)^{\log_3 12}=\left(3^{\log_3 12}\right)^2=12^2=144\).

\(\log _{2} 8=\log_22^3=3\log_22=3\).

\(\log _{3} \displaystyle\frac{1}{9}=-2\) vì \(3^{-2}=\displaystyle\frac{1}{9}\).

\(\log _3 81=4\) vì \(2^{4}=81\).

\(\log _{10} \displaystyle\frac{1}{100}=-2\) vì \(10^{-2}=\displaystyle\frac{1}{100}\).

\(\log _{5} \sqrt[3]{5}=\log _{5} 5^{\frac{1}{3}}=\displaystyle\frac{1}{3}\).

\(4^{\log _{2} 7}=\left(2^{2}\right)^{\log _{2} 7}=\left(2^{\log _{2} 7}\right)^{2}=7^{2}=49\).

\(\log _3 81=\log _3 3^4=4\).

\(36^{\log _{6} 8}=\left(6^{2}\right)^{\log _{6} 8}=\left(6^{\log _{6} 8}\right)^{2}=8^{2}=64\).

\(\log 0{,}0001=\log 10^{-4}=-4\).

\(\ln \mathrm{e}^{2}=2\).

\(\log_{\frac{1}{4}} 8=\log_{2^{-2}} 2^3=\displaystyle\frac{3}{-2}\log_2 2=-\displaystyle\frac{3}{2}\cdot 1=-\displaystyle\frac{3}{2}\).

\(\log \sqrt{1000}=\log 10^{\frac{3}{2}}=\displaystyle\frac{3}{2}\log 10=\displaystyle\frac{3}{2}\cdot 1=\displaystyle\frac{3}{2}\).

\(\log_9 27=\log_3^2 3^3=\displaystyle\frac{3}{2}\log_3 3=\displaystyle\frac{3}{2}\cdot 1=\displaystyle\frac{3}{2}\).

\(\log_5 \sqrt[3]{25}=\log_5 25^{\displaystyle\frac{1}{3}}=\displaystyle\frac{2}{3}\log_5 5=\displaystyle\frac{2}{3}\cdot 1=\displaystyle\frac{2}{3}\).

\(\log _{a} a^{-3}=-3\).

Ta có \(A=a^{\log_{a^{\frac{1}{2}}} 4}=a^{2\log_a 4}=(a^{\log_a 4})^2=4^2=16\).

Ta có \(P=a^{\log_{\sqrt{a}}3}=a^{2\log_a3}=\left(a^{\log_a3}\right)^2=3^2=9\).

Ta có \(A=\log^2_{a^2}a^4=\left(\log_{a^2}a^4\right)^2=\left(4\cdot \displaystyle\frac{1}{2}\log_aa\right)^2=4.\)

\(A=\log_a\displaystyle\frac{\sqrt{a^3}}{a\sqrt[4]{a}}=\log_a\displaystyle\frac{a^{\tfrac{3}{2}}}{a^{1+\tfrac{1}{4}}}=\log_aa^{\tfrac{3}{2}-\tfrac{5}{4}}=\displaystyle\frac{3}{2}-\displaystyle\frac{5}{4}=\displaystyle\frac{1}{4}\).

Ta có \(\log_a \left (a\sqrt[3]{a}\right ) = \log_a a^{\tfrac{4}{3}} = \displaystyle\frac{4}{3}\).

\(A=a^{2\log_{\sqrt{a}} 3}=a^{4\log_a 3}=a^{\log_a 3^4}=3^4\).

\(\log_3 \left(9^2\cdot 3^2\right)=\log_3 9^2+\log_3 3^2=2\log_3 3^2+2\log_3 3=2\cdot 2\log_3 3+2=4+2=6\).

\(\log_2 \displaystyle\frac{2}{3}+\log_2 12=\log_2 \left(\displaystyle\frac{2}{3}\cdot 12\right)=\log_2 2^3=3\log_2 2=3\cdot 1=3\).

\(\log_5 4+\log_5 \displaystyle\frac{1}{4}=\log_5 4+\log_5 4^{-1}=\log_5 4+(-1)\log_5 4=0\).

\(\log_2 28-\log_2 7=\log_2 \displaystyle\frac{28}{7}=\log_2 4=\log_2 2^2=2\log_2 2=2\cdot 1=2\).

\(\log _{6} 9+\log _{6} 4=\log _{6}(9\cdot 4)=\log _{6} 36=2\).

\(\log _{5} 100-\log _{5} 20=\log _{5} \displaystyle\frac{100}{20}=\log _{5} 5=1\).

\(\ln \left(\sqrt{5}+2\right)+\ln \left(\sqrt{5}-2\right)=\ln \left(\left(\sqrt{5}+2\right)\cdot \left(\sqrt{5}-2\right)\right)=\ln 1=0\).

\(\log 400-\log 4=\log\displaystyle\frac{400}{4}=\log 100=2\).

\(\log _{4} 8+\log _{4} 12+\log _{4} \displaystyle\frac{32}{3}=\log \left(8\cdot 12\cdot \displaystyle\frac{32}{3}\right)=\log_4 1024=\log_4 4^5=5\).

\(\log _4 2+\log _4 32=\log _4(2 \cdot 32)=\log _4 64=\log _4 4^3=3 \log _4 4=3\).

\(\log _2 80-\log _2 5=\log _2 \displaystyle\frac{80}{5}=\log _2 16=\log _2 2^4=4 \log _2 2=4\).

\(\log _8 16-\log _8 2=\log_82+\log_88-\log_82=1\).

\(\log _{5} 15-2 \log _{5} \sqrt{3}=\log _{5} 15-\log _{5}(\sqrt{3})^{2}=\log _{5} 15-\log _{5} 3=\log _{5} \displaystyle\frac{15}{3}=\log _{5} 5=1\).

Ta có:

\(\log_{6} 9+\log_6 4=\log_6 (9\cdot 4)=\log_6 36=\log_6 6^2=2\log_6 6=2\cdot 1=2\).

Ta có:

\(\log_5 2-\log_5 50=\log_5 \displaystyle\frac{2}{50}=\log_5 \displaystyle\frac{1}{25}=\log_5 5^{-2}=-2\cdot \log_5 5=-2\cdot 1=-2\).

Ta có:

\(\log_3 \sqrt{5}-\displaystyle\frac{1}{2}\log_3 15=\log_3 \displaystyle\frac{\sqrt{5}}{\sqrt{15}}=\log_3 \displaystyle\frac{1}{\sqrt{3}}=\log_3 1-\log_3 3^{\frac{1}{2}}=0-\displaystyle\frac{1}{2}\log_3 3=0-\displaystyle\frac{1}{2}\cdot 1=-\displaystyle\frac{1}{2}\).

\(P=\log_3 5^2-\log _3 50+\log _3 36^{\frac{1}{2}}=\log_3 25-\log _3 50+\log _3 6=\log_3\displaystyle\frac{25\cdot 6}{50}=\log_3 3=1\).

\(\log _2 6 \cdot \log _6 8=\log_28=\log_22^3=3\log_22=3\).

\(\log_2 3\cdot \log_3 \displaystyle\frac{1}{4}=\log_3 2\cdot \displaystyle\frac{\log_2 \displaystyle\frac{1}{4}}{\log_2 3}=\log_2 2^{-2}=-2\log_2 2=-2\cdot 1=-2\).

\(\log_4 5\cdot \log_5 6\cdot \log_6 8=\log_4 5\cdot \displaystyle\frac{\log_4 6}{\log_4 5} \cdot \displaystyle\frac{\log_4 8}{\log_4 6}=\log_4 8=\log_{2^2} (2^3)=\displaystyle\frac{3}{2}\log_2 2=\displaystyle\frac{3}{2}\).

Ta có:

\(\log_{2} 9 \cdot \log_3 4=\log_2 9 \cdot \displaystyle\frac{\log_2 4}{\log_2 3}=\log_2 3^2 \cdot \displaystyle\frac{2\log_2 2}{\log_2 3}=2\log_2 3 \cdot \displaystyle\frac{2\cdot 1}{\log_2 3}=2\cdot 2=4\).

Ta có:

\(\log_2 3 \cdot \log_9 \sqrt{5}\cdot \log_5 4=\log_2 3\cdot \displaystyle\frac{\log_2 \sqrt{5}}{\log_2 9}\cdot \log_5 2^2 =\log_2 3\cdot \displaystyle\frac{1}{2}\cdot \displaystyle\frac{\log_2 5}{2\log_2 3}\cdot 2\log_5 2=\displaystyle\frac{1}{2}\log_2 5\cdot \log_5 2=\displaystyle\frac{1}{2}\).

\(\log _{a}\left(a^{2} b^{3}\right)=\log _{a}a^2+\log _{a}b^3=2+3\log _{a}b=8\).

\(\log _{a} \displaystyle\frac{a \sqrt{a}}{b \sqrt[3]{b}}=\log_a a\sqrt{a}-\log_ab\sqrt{b}=\log_aa^\frac{3}{2}-\log_ab^\frac{4}{3}=\displaystyle\frac{3}{2}-\displaystyle\frac{4}{3}\log_ab=-\displaystyle\frac{7}{6}\).

Ta có \(\log _{12}18=\displaystyle\frac{\log _218}{\log _212}=\displaystyle\frac{\log _29+\log _22}{\log _24+\log _23}=\displaystyle\frac{2\log _23+1}{2+\log _23}=\displaystyle\frac{2a+1}{2+a}\).

Ta có \(a=\log_{20}5=\log_{20}\displaystyle\frac{20}{4}=1-2\log _{20}2\).

Suy ra \(\log _{20}2=\displaystyle\frac{1-a}{2}\) hay \(\log _{2}{20}=\displaystyle\frac{2}{1-a}\).

Ta có \(\log \sqrt[3]{\displaystyle\frac{8}{5}}=\displaystyle\frac{1}{3}\log \displaystyle\frac{16}{10}=\displaystyle\frac{1}{3}\left(4\log 2-1\right)=\displaystyle\frac{4a-1}{3}\).

Ta có \(\log_8{25}=\log_{2^3}{5^2}=\displaystyle\frac{2}{3}\log_2{5}=\displaystyle\frac{2}{3}a\).

Ta có \(\log_2x = \displaystyle\frac{1}{2} \Leftrightarrow x = \sqrt{2} \Rightarrow x^2 = 2\).

Khi đó

\(P=\displaystyle\frac{\log_2(4x)+\log_2\displaystyle\frac{x}{2}}{x^2-\log_{\sqrt{2}}x}=\displaystyle\frac{\log_2\left(2x^2\right)}{x^2-2\log_2x}=\displaystyle\frac{1+2\log_2x}{x^2-2\log_2x}=\displaystyle\frac{1+2\cdot \displaystyle\frac{1}{2}}{2-2\cdot \displaystyle\frac{1}{2}}=2.\)

\(\log _{a}(2 b)+\log _{a}\left(\displaystyle\frac{b^{2}}{2}\right)=\log_a2+\log_ab+\log_ab^2-\log_a2=\log_ab+2\log_ab=3\log_ab=6\).

\(\log_4\sqrt{a}=5 \Leftrightarrow \displaystyle\frac{1}{4} \log_2 a = 5 \Leftrightarrow \log_2 a = 20.\)

Từ \(\log_6 a=2\) (\(a>0\)) \(\Leftrightarrow a=6^2\).

Khi đó \(I=\log_6 \left(\displaystyle\frac{1}{a}\right)=\log_6\left(6^2\right)^{-1}=-2\).

Ta có

\begin{align*}\log_{10}250=\ &\dfrac{\log_2 250}{\log_2 10}=\dfrac{\log_2(2\cdot5^3)}{\log_2(2\cdot5)}=\dfrac{1+3\log_25}{1+\log_25}=\dfrac{1+3a}{1+a}.\end{align*}

Ta có

\[\log_{100}2\sqrt{5}=\dfrac{\log_2 2\sqrt{5}}{\log_2 100}=\dfrac{1+\frac{1}{2}\log_25}{2+2\log_25}=\dfrac{1+0{,}5a}{2+2a}.\]

Ta có

\[\log_4 18=\dfrac{\log_3 18}{\log_3 4}=\dfrac{\log_3(2\cdot3^2)}{\log_32^2}=\dfrac{\log_32+2}{2\log_32}=\dfrac{a+2}{2a}.\]

Ta có

\[\log_4 50=\dfrac{\log_5 50}{\log_5 4}=\dfrac{\log_52+2}{2\log_52}=\dfrac{b+2}{2b}.\]

Từ $\log_3 18=a\Rightarrow \log_3(2\cdot3^2)=a\Rightarrow \log_32+2=a\Rightarrow \log_32=a-2$.\\

Ta có

\[\log_4 72=\dfrac{\log_3 72}{\log_3 4}=\dfrac{\log_3(2^3\cdot3^2)}{\log_32^2}=\dfrac{3\log_32+2}{2\log_32}=\dfrac{3(a-2)+2}{2(a-2)}=\dfrac{3a-4}{2a-4}.\]

Từ $\log_29=a\Rightarrow 2\log_23=a\Rightarrow \log_23=\dfrac{a}{2}$.\\

Ta có

\begin{align*}\log_{6}81=\ &\dfrac{\log_2 81}{\log_2 6}=\dfrac{\log_2 3^4}{\log_2(2\cdot3)}=\dfrac{4\log_23}{1+\log_23}=\dfrac{2a}{1+0{,}5a}.\end{align*}

Từ $\log_36=a\Rightarrow \log_3(2\cdot3)=a\Rightarrow \log_32=a-1$.\\

Ta có

\[\log_{12} 18=\dfrac{\log_3 18}{\log_3 12}=\dfrac{\log_3(2\cdot3^2)}{\log_3(2^2\cdot3)}=\dfrac{\log_32+2}{2\log_32+1}=\dfrac{a-1+2}{2(a-1)+1}=\dfrac{a+1}{2a-1}.\]

Ta có

\begin{align*}\log_{12}90=\ &\dfrac{\log_2 90}{\log_2 12}=\dfrac{\log_2(2\cdot3^2\cdot5)}{\log_2(2^2\cdot3)}=\dfrac{1+2\log_23+\log_25}{2+\log_23}=\dfrac{1+2a+b}{2+a}.\end{align*}

Ta có

\begin{align*}\log_{15}20=\ &\dfrac{\log_2 20}{\log_2 15}=\dfrac{\log_2(2^2\cdot5)}{\log_2(3\cdot5)}=\dfrac{2+\log_25}{\log_23+\log_25}=\dfrac{2+b}{a+b}.\end{align*}

Ta có

\begin{align*}\log_{30}54=\ &\dfrac{\log_2 54}{\log_2 30}=\dfrac{\log_2(3^2\cdot5)}{\log_2(2\cdot3\cdot5)}=\dfrac{2\log_23+\log_25}{1+\log_23+\log_25}=\dfrac{2a+b}{1+a+b}.\end{align*}

Ta có \(\heva{& \log_a x =2 \\ & \log_b x =3} \Leftrightarrow \heva{& x=a^2 \\ & x = b^3} \Rightarrow a^2=b^3 \Leftrightarrow a = b^{\frac{3}{2}}\).

Do đó, \(P=\log_{\frac{a}{b^2}}x=\log_{b^{-\frac{1}{2}}}x=-2 \log_b x = -2 \times 3=-6\).

\(\log \left(a^2 \cdot b^3\right)=2\log a+3\log b =2 \cdot2 + 3\cdot 3 =13\).

Theo tính chất lôgarit, suy ra \(P= \log_a (b^2c^3) =\log_a b^2 +\log_a c^3 =2 \log_a b + 3 \log_a c=13\).

Ta có \(P=\log_2\left(a^2b^3\right)=\log_2 a^2+\log_2 b^3=2\log_2 a+3\log_2 b=2x+3y\).

Ta có

\(P=\log_a\left(\displaystyle\frac{b^2}{c^3}\right)\Leftrightarrow P=2\log_ab-3\log_ac\Leftrightarrow P=2\cdot 2-3\cdot 3=-5.\)

Ta có \(\log_{ab} b =3\Leftrightarrow \log_{b} (ab) =\displaystyle\frac{1}{3}\Leftrightarrow \log_b a +1 =\displaystyle\frac{1}{3}\Leftrightarrow \log_b a =-\displaystyle\frac{2}{3}\).

Do đó

\begin{align*}\log_{\sqrt{ab}} \left(\displaystyle\frac{a}{b^2}\right)=\ &2\log_{ab} \left( \displaystyle\frac{a}{b^2} \right)\\ =\ &2(\log_{ab} a-2\log_{ab} b)\\ =\ &2(\log_{ab} b\cdot \log_b a -2\cdot 3)\\ =\ &2\left( 3\cdot \left( -\displaystyle\frac{2}{3} \right) -6 \right)=-16.\end{align*}

Ta có

\(\log_3\left(\log_9x\right)=0\Leftrightarrow \log_9x=1\Leftrightarrow x=9.\)

Từ đó \((\log_3x)^2=(\log_39)^2=4\).

Ta có:

\(\log_4 9=\displaystyle\frac{\log 9}{\log 4}=\displaystyle\frac{\log 3^2}{\log 2^2}=\displaystyle\frac{2\log 3}{2\log 2}=\displaystyle\frac{b}{a}\).

Ta có:

\(\log_6 12=\displaystyle\frac{\log 12}{\log 6}=\displaystyle\frac{2\log 2+\log 3}{\log 2+\log 3}=\displaystyle\frac{2a+b}{a+b}\).

Ta có \(\log_5 6=\displaystyle\frac{\log 6}{\log 5}=\displaystyle\frac{\log 2+\log 3}{\log (2+3)}\).

Lại có \(\log 2=a\Rightarrow 10^a=2\) và \(\log 3=b\Rightarrow 10^b=3\). Do đó

\(\log_5 6=\displaystyle\frac{\log 6}{\log 5}=\displaystyle\frac{\log 2+\log 3}{\log 10 -\log 2}=\displaystyle\frac{a+b}{1-a}.\)

\(\log_9 10=\displaystyle\frac{\log_2 10}{\log_2 9}=\displaystyle\frac{\log_2 (2\cdot 5)}{\log_2 3^2}=\displaystyle\frac{\log_2 2+\log_2 5}{2\log_2 3}=\displaystyle\frac{1+b}{2a}\).

\(\log_{12} 21=\displaystyle\frac{\log_3 21}{\log_3 12}=\displaystyle\frac{\log_3 3+\log_3 7}{\log_3 3+\log_3 4}=\displaystyle\frac{1+b}{1+2a}\).

Ta có \(\log_{a^2}\left(\sqrt[3]{b^5c^4}\right)=\displaystyle\frac{1}{6}\log_a(b^5c^4)=\displaystyle\frac{5}{6}\log_ab+\displaystyle\frac{2}{3}\log_ac=\displaystyle\frac{5}{6\log_ba}+\displaystyle\frac{2\log_bc}{3\log_ba}=\displaystyle\frac{5+4y}{6x}\).

Ta có

\begin{align*}\log_2\sqrt[6]{360}=\ &\displaystyle\frac{1}{6}\log_2(2^3\cdot 3^2\cdot 5)=\displaystyle\frac{1}{6}\left(3\log_22+2\log_23+\log_25\right)\\=\ &\displaystyle\frac{1}{2}+\displaystyle\frac{1}{3}\log_23+\displaystyle\frac{1}{6}\log_25=\displaystyle\frac{1}{2}+\displaystyle\frac{1}{3}a+\displaystyle\frac{1}{6}b.\end{align*}

Ta có

\(\log_2\displaystyle\frac{40}{3}=\log_240-\log_23=\log_2(5\cdot 8)-\displaystyle\frac{1}{2}\log_2 9=\log_25+\log_28-\displaystyle\frac{1}{2}\log_29=a+3-\displaystyle\frac{1}{2}b.\)

Ta có

\(\log_{15}20 =\displaystyle\frac{\log_220}{\log_215}=\displaystyle\frac{2+\log_25}{\log_23+\log_25}=\displaystyle\frac{2+a}{\displaystyle\frac{1}{b}+a}=\displaystyle\frac{2b+ab}{1+ab}.\)

Ta có \(\log_{10}6=\displaystyle\frac{\log_5 6}{\log_5 10}=\displaystyle\frac{\log_5 2+\log_5 3}{1+\log_5 2}=\displaystyle\frac{a+b}{1+a}\).

Ta có \(\log_65 =\displaystyle\frac{1}{\log_56}= \displaystyle\frac{1}{\log_52+\log_53}=\displaystyle\frac{1}{\displaystyle\frac{1}{\log_25}+\displaystyle\frac{1}{\log_35}}=\displaystyle\frac{1}{\displaystyle\frac{1}{a}+\displaystyle\frac{1}{b}}= \displaystyle\frac{ab}{a+b}.\)

Ta có \(\log_{10}6=\displaystyle\frac{\log_5 6}{\log_5 10}=\displaystyle\frac{\log_5 2+\log_5 3}{1+\log_5 2}=\displaystyle\frac{a+b}{1+a}\).

Ta có \(\log_65 =\displaystyle\frac{1}{\log_56}= \displaystyle\frac{1}{\log_52+\log_53}=\displaystyle\frac{1}{\displaystyle\frac{1}{\log_25}+\displaystyle\frac{1}{\log_35}}=\displaystyle\frac{1}{\displaystyle\frac{1}{a}+\displaystyle\frac{1}{b}}= \displaystyle\frac{ab}{a+b}.\)

\(\log_a b^2+\log_{a^3} b^8=2\log_a b+\displaystyle\frac{8}{3}\log_{a} b =\displaystyle\frac{14}{3} \log_a b\).

Ta có \(P=2\log_2 a-\log_{\frac{1}{2}} b^2=\log_2 a^2+\log_2 b^2=\log_2 a^2b^2=\log_2 \left(ab\right)^2\).

Ta có \(A=\log_2\displaystyle\frac{1}{2^a}+\log_2\displaystyle\frac{1}{2^b} =\log_2 2^{-a}+\log_2 2^{-b} = -a-b.\)

\(\log_{a^3}\left(ab^6\right)=\displaystyle\frac{1}{3}\log_{a}\left(ab^6\right)=\displaystyle\frac{1}{3}\left(\log_{a}a+\log_{a}b^6\right)=\displaystyle\frac{1}{3}+2\log_a{b}\).

\(\log_{a^4}b^2=\displaystyle\frac{1}{4}\cdot 2 \log_a(-b)=\displaystyle\frac{1}{2}\log_a(-b)\) (vì \(b<0\)).

Ta có: \(a^{3} b^{2}=100\).

Suy ra \(\log a^3b^2=\log 100\Leftrightarrow\log a^3+\log b^2=\log 10^2\Leftrightarrow3\log a+2\log b=2\).

Vậy \(P=2\).

\(A=\ln \left(\displaystyle\frac{x}{x-1}\right)+\ln \left(\displaystyle\frac{x+1}{x}\right)-\ln \left(x^2-1\right)=\ln\left[\displaystyle\frac{\displaystyle\frac{x}{x-1}\cdot\displaystyle\frac{x+1}{x}}{x^2-1}\right]\).

Hay \(A=\ln\left[\displaystyle\frac{x+1}{(x+1)(x-1)^2}\right]=-2\ln(x-1)\).

\(B=21 \log _3 \sqrt[3]{x}+\log _3\left(9 x^2\right)-\log _3 9=\log_3\displaystyle\frac{\left(\sqrt[3]{x}\right)^{21}\cdot9x^2}{9}=9\log_3x\).

\(A=\log _{\displaystyle\frac{1}{3}} 5+2 \log _9 25-\log _{\sqrt{3}} \displaystyle\frac{1}{5}=-\log_35+2\log_35+2\log_35=3\log_35\).

\(B=\log _{a} M^2+\log _{a^2} M^4=2\log_aM+\displaystyle\frac{4}{2}\log_aM=4\log_aM\).

Áp dụng công thức \(\log_ab\cdot\log_bc=\log_ac\), ta được

\(A=\log _2 3 \cdot \log _3 4 \cdot \log _4 5 \cdot \log _5 6 \cdot \log _6 7 \cdot \log _7 8=\log_28=\log_22^3=3\).

\(B=\log _2 2 \cdot \log _2 4 \cdots \log _2 2^n=\log _2 2^1 \cdot \log _2 2^2 \cdots \log _2 2^n=1\cdot2\cdots n=n!\).

Đặt \(t=\log_ab\). Ta có

\begin{align*}&\log_{a}^2b-8\log_b(a\sqrt[3]{b})=-\displaystyle\frac{8}{3}\\ \Leftrightarrow\ &\log_{a}^2b-8\left(\log_b(a)+\displaystyle\frac{1}{3}\right)=-\displaystyle\frac{8}{3}\\ \Leftrightarrow\ & \log_{a}^2b-\displaystyle\frac{8}{\log_ab}=0\\ \Leftrightarrow\ &\log_ab=2.\end{align*}

Mặt khác \(P=\log_a\left(a\sqrt[3]{ab}\right)=\displaystyle\frac{4}{3}+\displaystyle\frac{1}{3}\log_{a}b=2.\)

Ta có: \(\log_ac>0 \Rightarrow c>0\) và \(c\ne 1\). Suy ra

\(\log_{ab}c=\displaystyle\frac{1}{\log_cab}=\displaystyle\frac{1}{\log_ca+\log_cb}=\displaystyle\frac{1}{\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}}=\displaystyle\frac{xy}{x+y}.\)

Ta có \(\log_5x=\log_{\sqrt[3]{5}}b-2\log_{\frac{1}{5}}a\Leftrightarrow \log_5x=3\log_5b+2\log_5a\)

\(\Leftrightarrow \log_5x=\log_5\left(a^2b^3\right)\Leftrightarrow x=a^2b^3\).

Ta có: \(3\log a+2\log b=1 \Leftrightarrow \log a^{3}+\log b^{2}=1\Leftrightarrow \log\left(a^{3}b^{2}\right)=1\Leftrightarrow a^{3}b^{2}=10.\)

Đẳng thức ở giả thiết tương đương với \(\left(\displaystyle\frac{a+b}3\right)^2=ab.\)

Lấy lô-ga-rít cơ số 2 vào 2 vế của đẳng thức này, ta được \(2\log_2\displaystyle\frac{a+b}3=\log_2a+\log_2b.\)

Ta có \(\log_4a=\log_9b=\log_6(a-b)=t\Rightarrow a=4^t;b=9^t;a-b=6^t\)

\begin{align*}\Rightarrow\ & 4^t-9^t=6^t\\\Leftrightarrow\ & {\left(\displaystyle\frac{2}{3}\right)}^{2t}-{\left(\displaystyle\frac{2}{3}\right)}^t-1=0\\\Rightarrow\ &\hoac{& {\left(\displaystyle\frac{2}{3}\right)}^t=\displaystyle\frac{1-\sqrt{5}}{2}<0\quad \text{(loại)}\\ &{\left(\displaystyle\frac{2}{3}\right)}^t=\displaystyle\frac{1+\sqrt{5}}{2}}\\\Rightarrow\ & M=\displaystyle\frac{5+\sqrt{5}}{10}.\end{align*}

Ta có \(9a^2+b^2=10ab\Leftrightarrow 9a^2+6ab+b^2=16ab\Leftrightarrow (3a+b)^2=16ab\).

\(\Leftrightarrow \log\displaystyle\frac{(3a+b)^2}{16}=\log (ab)\Leftrightarrow 2\log\displaystyle\frac{3a+b}{4}=\log a+\log b\Leftrightarrow \log\displaystyle\frac{3a+b}{4}=\displaystyle\frac{1}{2}\left(\log a+\log b\right)\).

Ta thấy \(\log_b a = \log_a b + 2 \Leftrightarrow \log^2_a b + 2\log_a b - 1 = 0 \Rightarrow \log_a b = - 1 + \sqrt{2}\) (vì \(a, b > 1\)).

Ta có

\(a^2+b^2=7ab \Leftrightarrow (a+b)^2=9ab\Leftrightarrow \left(\displaystyle\frac{a+b}{3}\right)^2=ab.\)

Suy ra

\(\log_7\left(\displaystyle\frac{a+b}{3}\right)^2=\log_7(ab)\Leftrightarrow \log_7\displaystyle\frac{a+b}{3}=\displaystyle\frac{1}{2}\left(\log_7a+\log_7b\right).\)

Ta có \(\log 2\cdot \log_2 a-\log b=2\Leftrightarrow \log \displaystyle\frac{a}{b}=2\Leftrightarrow \displaystyle\frac{a}{b}=100.\)

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Với mọi số thực dương \(a\) và \(b\), ta có \(a^2 + b^2 = 2ab \Leftrightarrow (a+b)^2 = 4ab\).

Do đó, ta có

\(\log_2(a+b)^2 = \log_2(4ab) \Leftrightarrow 2\log_2(a+b) = \log_24+\log_2a+\log_2b\) \(\Leftrightarrow \log_2(a+b)=\displaystyle\frac{1}{2}(2+\log_2a+\log_2b)\).

Từ giả thiết ta có

\(\log_2\left( \displaystyle\frac{b^2}{a^3}\right) = 2 \Leftrightarrow b^2 = 4a^3\).

Ta có \(a^2+b^2=7ab\Leftrightarrow (a+b)^2=9ab.\) Do đó

\(\log_2 (a+b)^2=\log_2 (9ab)\Leftrightarrow 2\log_2(a+b)=2\log_2 3+\log_2 a+\log_2 b\Leftrightarrow 2\log_2\displaystyle\frac{a+b}{3}=\log_2 a+\log_2 b.\)

Ta có \(a^2+b^2=23ab\Leftrightarrow (a+b)^2=25ab\Leftrightarrow \left(\displaystyle\frac{a+b}{5}\right)^2=ab\)\;(*)

Lấy lôgarit cơ số \(5\) hai vế của (*) ta được

\(2\left[\log_{5}(a+b)-\log_{5}5\right]=\log_5(ab)\Leftrightarrow 2\left[\log_{5}(a+b)-1\right]=\log_{5}a+\log_{5}b\)

\(\Leftrightarrow \log_5(a+b)=\displaystyle\frac{1}{2}\left(\log_5{a}+\log_5{b}\right)+1=1+\log_{25}a+\log_{25}b\).

Ta có \(a^2+b^2=23ab\Leftrightarrow (a+b)^2=25ab\Leftrightarrow \left(\displaystyle\frac{a+b}{5}\right)^2=ab\)\;(*)

Lấy lôgarit tự nhiên hai vế của (*), ta được

\(2\ln\displaystyle\frac{a+b}{5}=\ln(ab)=\ln a+\ln b\Rightarrow \ln\displaystyle\frac{a+b}{5}=\displaystyle\frac{\ln a+\ln b}{2}\).

Ta có \(a^2+b^2=23ab\Leftrightarrow (a+b)^2=25ab\Leftrightarrow \left(\displaystyle\frac{a+b}{5}\right)^2=ab\)\;(*)

Lấy lôgarit thập phân hai vế của (*), ta có

\(2\log\displaystyle\frac{a+b}{5}=\log a+\log b\).

Từ giả thiết ta có \(\log_2\left(\displaystyle\frac{b^2}{a^3}\right)=2\Leftrightarrow b^2=4a^3\).

a) Khi \(\left[\mathrm{H}^{+}\right]=0{,}01\), ta có: \(\mathrm{pH}=-\log 0,01=-\log 10^{-2}=2\).

b) Nồng độ ion hydrogen trong dung dịch đó là \(\left[\mathrm{H}^{+}\right]=10^{-7{,}4}\).

Ta có: \(r=6 \%=0{,}06\). Do đó thời gian cần thiết để tăng gấp đôi khoản đầu tư là

\(t=\displaystyle\frac{\ln 2}{r}=\displaystyle\frac{\ln 2}{0{,}06} \approx 11{,}6\, (\text {năm}).\)

Ta có: \(A=100 \cdot(1+0{,}06)^n=100 \cdot 1{,}06^n\).

Với \(A=150\), ta có: \(100 \cdot 1{,}06^n=150\) hay \(1{,}06^n=1{,}5\), tức là \(n=\log _{1{,}06} 1{,}5 \approx 6{,}96\).

Vì gửi tiết kiệm kì hạn \(12\) tháng (tức là \(1\) năm) nên \(n\) phải là số nguyên.\\ Do đó ta chọn \(n=7\).

Vậy sau ít nhất \(7\) năm thì bác An nhận được số tiền không dưới \(150\) triệu đồng.

Đỉnh Everest có độ cao \(8850 \mathrm{~m}\) so với mực nước biển suy ra

\(8850=15500(5-\log p)\Leftrightarrow \log p=\displaystyle\frac{1373}{310}\Leftrightarrow p=26855{,}44\mathrm{~(pascal)}.\)

Áp suất không khí ở đỉnh Everest là \(p=26855{,}44\mathrm{~(pascal)}\).

a) Cuộc trò chuyện bình thường có cường độ \(I=10^{-7} \mathrm{~W} / \mathrm{m}^2\) có mức cường độ âm là

\(L(I)=10 \log \displaystyle\frac{10^{-7}}{10^{-12}}=50~(\mathrm{dB}).\)

b) Giao thông thành phố đông đúc có cường độ \(I=10^{-3} \mathrm{~W} / \mathrm{m}^2\)

có mức cường độ âm là

\(L(I)=10 \log \displaystyle\frac{10^{-3}}{10^{-12}}=90~(\mathrm{dB}).\)

Độ pH của đầm là: \(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left( 8\cdot 10^{-8}\right)\approx 7{,}1\).

Do vậy, độ pH của đầm không thích hợp cho tôm sú phát triển.

Số lần phân chia: \(N=N_0 \cdot 2^n \Rightarrow n=\log_2\left(\displaystyle\frac{N}{N_0}\right)=\log_2\left(\displaystyle\frac{6 \cdot 10^{27}}{5 \cdot 10^{-13}}\right)=\log_2\left(1{,}2\cdot 10^{40}\right)\approx 133\).

Thời gian cần thiết là: \(133: 3\approx 44{,}3\) giờ.

Vậy sau 45 giờ thì khối lượng do tế bào vi khuẩn này sinh ra sẽ đạt tới khối lượng của Trái Đất.

a) Độ lớn của trận động đất có biên độ \(A\) là

i) \(10^{5{,}1}A_0 \Rightarrow M=\log \displaystyle\frac{10^{5{,}1}A_0}{A_0}=5{,}1\).

ii) \(65~000 A_0\Rightarrow M=\log \displaystyle\frac{65~000 A_0}{A_0} =\log (65\cdot 10^3)\approx 4{,}81\).

b) Gọi \(M_N\) và \(M_P\) lần lượt là độ lớn của các trận động đất tại địa điểm \(N\) và \(P\).

Gọi \(A\) là biên độ lớn nhất ghi được bởi máy đo địa chấn tại địa điểm \(P\). Ta có

\(M_P=\log \displaystyle\frac{A}{A_0};\quad M_N=\log \displaystyle\frac{3A}{A_0}=\log 3+\log \displaystyle\frac{A}{A_0}.\)

Do \(\log 3\approx 0{,}3>0\) nên \(M_N>M_P\).

a) \(\text{pH}=-\log 0{,}0001=-\log 10^{-4} =4\log 10=4\).

Do \(4<7\) nên dung dịch có tính acid.

b) Kí hiệu pH\(_A\), pH\(_B\) lần lượt là độ pH của hai dung dịch \(A\) và \(B\). \([\text{H}^+]_A\), \([\text{H}^+]_B\) lần lượt là nồng độ của hai dung dịch \(A\) và \(B\). Ta có

\(\text{pH}_A=-\log[\text{H}^+]_A=-\log\left(2[\text{H}^+]_B\right)=-\log 2-\log[\text{H}^+]_B=-\log 2+\text{pH}_B.\)

Suy ra \(\text{pH}_B-\text{pH}_A=\log 2\approx 0{,}301\).

Vậy dung dịch \(B\) có độ pH lớn hơn và lớn hơn khoảng \(0{,}301\).