Pojęcie logarytmu
\mathrm{\log _{\mathrm{2}}8=}\textcolor{#005FD4}{\mathrm{3}} log2 8 = 3 , ponieważ \left.\mleft.2\mright.\right.^{\mleft.\textcolor{#005FD4}{3}\mright.}=8 23 = 8
\mathrm{\log _{\mathrm{5}}1=}\textcolor{#005FD4}{\mathrm{0}} log5 1 = 0 , ponieważ \left.\mleft.5\mright.\right.^{\mleft.\textcolor{#005FD4}{0}\mright.}=1 50 = 1
\mathrm{\log _2} log2 \frac{1}{8}=\textcolor{#005FD4}{-3} 18 = −3 , ponieważ\left.\mleft.2\mright.\right.^{\mleft.\textcolor{#005FD4}{-3}\mright.}=\frac{1}{8} 2−3 = 18
\mathrm{\log _{\mathrm{49}}7=} log49 7 = \frac{\textcolor{#005FD4}{1}}{\textcolor{#005FD4}{2}} 12 , ponieważ\left.\mleft.49\mright.\right.^{\mleft.\frac{\textcolor{#005FD4}{1}}{\textcolor{#005FD4}{2}}\mright.}=7 4912 = 7
\mathrm{\log _{}100=}\textcolor{#005FD4}{\mathrm{2}} log 100 = 2 , ponieważ \left.\mleft.10\mright.\right.^{\mleft.\textcolor{#005FD4}{2}\mright.}=100 102 = 100
\mathrm{\log _{\mathrm{a}}b=} loga b = x x , gdy \left.\mleft.a\mright.\right.^{\mleft.x\mright.}=\mathit{b} ax = b
