Nếu $a > 1$ và $b > c > 0$ thì:

$\left( {1;2} \right) \cup \left( {3; + \infty } \right)$

$\left( { - \infty ;1} \right) \cup \left( {2;3} \right)$

$\left( {1;2} \right) \cap \left( {3; + \infty } \right)$

$\left( { - \infty ;1} \right) \cap \left( {2;3} \right)$

$x = 63$

$x = 65$

$x = 82$ 

$x = 80$

\({x^{\frac{2}{3}}} + 5 = 0\)

${(3x)^{\frac{1}{3}}} + {\left( {x - 4} \right)^{\frac{2}{5}}} = 0$

$\sqrt {4x - 8}  + 2 = 0$

$2{x^{\frac{1}{2}}} - 3 = 0$

$0 < a < 1$

$a > 0$

$a > 1$

$a < 0$

\(\alpha  = 0\)

\(\alpha  = 1\)

\(\alpha  > 1\)

\(0 < \alpha  < 1\) 

$\ln {\left( {ab} \right)^2} = \ln \left( {{a^2}} \right) + \ln \left( {{b^2}} \right)$

$\ln \left( {\sqrt {ab} } \right) = \dfrac{1}{2}\left( {\ln a + \ln b} \right)$

$\ln \left( {\dfrac{a}{b}} \right) = \ln \left| a \right| - \ln \left| b \right|$

$\ln {\left( {\dfrac{a}{b}} \right)^2} = \ln \left( {{a^2}} \right) - \ln \left( {{b^2}} \right)$

\(T = A.{e^{Nr}}\)                 

\(T = N.{e^{Ar}}\)     

\(T = r.{e^{NA}}\)     

\(T = A.{e^{N - r}}\)