Przykładowe rozwiązanie:

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a) Dla

$f\left(x\right)={\log }_{0,1}x$  

mamy

$f\left(1\right)=0\to \left(1,0\right)$  

$f\left(10\right)=-1\to \left(10,-1\right)$  

$f\left(\frac{1}{10}\right)=1\to \left(\frac{1}{10},1\right)$  

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b) Dla

$f\left(x\right)={\log }_{0,4}x$  

mamy

$f\left(1\right)=0\to \left(1,0\right)$  

$f\left(\frac{5}{2}\right)=-1\to \left(\frac{5}{2},-1\right)$  

$f\left(\frac{2}{5}\right)=1\to \left(\frac{2}{5},1\right)$  

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c) Dla

$f\left(x\right)={\log }_{0,6}x$  

mamy

$f\left(1\right)=0\to \left(1,0\right)$  

$f\left(\frac{5}{3}\right)=-1\to \left(\frac{5}{3},-1\right)$  

$f\left(\frac{3}{5}\right)=1\to \left(\frac{3}{5},1\right)$  

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d) Dla

$f\left(x\right)={\log }_{2\sqrt{2}}x$  

mamy

$f\left(1\right)=0\to \left(1,0\right)$  

$f\left(2\right)={\left(2\sqrt{2}\right)}^2=8\to \left(2,8\right)$  

$f\left(4\right)={\left(2\sqrt{2}\right)}^4=64\to \left(2,64\right)$  

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e) Dla

$f\left(x\right)={\log }_{3\sqrt{5}}x$  

mamy

$f\left(1\right)=0\to \left(1,0\right)$  

$f\left(2\right)={\left(3\sqrt{5}\right)}^2=45\to \left(2,45\right)$  

$f\left(4\right)={\left(3\sqrt{5}\right)}^4=2025\to \left(4,2025\right)$  

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f) Dla

$f\left(x\right)={\log }_{2,5}x$  

mamy

$f\left(1\right)=0\to \left(1,0\right)$  

$f\left(\frac{2}{5}\right)=-1\to \left(\frac{2}{5},-1\right)$  

$f\left(\frac{5}{2}\right)=1\to \left(\frac{5}{2},1\right)$