$\text{a)}$  

${\log }_{x}16=4$  

Z definicji logarytmu:

$x^4=16$  

$x^4=2^4$  

Porównujemy podstawy potęg:

$x=2$   

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$\text{b)}$  

${\log }_{x}0,09=\frac{2}{3}$  

Z definicji logarytmu:

$x^{\frac{2}{3}}=0,09$  

$x^{\frac{2}{3}}=\frac{9}{100}$  

$x^{\frac{2}{3}}={\left(\frac{3}{10}\right)}^2$  

${\left(x^{\frac{2}{3}}\right)}^{\frac{3}{2}}={\left({\left(\frac{3}{10}\right)}^2\right)}^{\frac{3}{2}}$  

$x^{\frac{2}{3}\cdot \frac{3}{2}}={\left(\frac{3}{10}\right)}^{2\cdot \frac{3}{2}}$  

$x={\left(\frac{3}{10}\right)}^3$  

$x=\frac{27}{1000}$  

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$\text{c)}$  

${\log }_x\left(\frac{1}{16}\right)=-\frac{4}{5}$  

Z definicji logarytmu:

$x^{-\frac{4}{5}}=\frac{1}{16}$  

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

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

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

$x={\left(\frac{1}{2}\right)}^{-5}$  

$x=2^5$  

$x=32$   

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$\text{d)}$  

${\log }_x\sqrt{5}=\frac{1}{4}$  

Z definicji logarytmu:

$x^{\frac{1}{4}}=\sqrt{5}$  

$x^{\frac{1}{4}}=5^{\frac{1}{2}}$  

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

$x^{\frac{1}{4}\cdot 4}=5^{\frac{1}{2}\cdot 4}$  

$x=5^2$  

$x=25$   

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$\text{e)}$  

${\log }_{x}2\sqrt{8}=5$   

Z definicji logarytmu:

$x^5=2\sqrt{8}$  

$x^5=2\cdot 8^{\frac{1}{2}}$  

$x^5=2\cdot {\left(2^3\right)}^{\frac{1}{2}}$  

$x^5=2\cdot 2^{3\cdot \frac{1}{2}}$  

$x^5=2\cdot 2^{\frac{3}{2}}$  

$x^5=2^{\frac{5}{2}}$  

${\left(x^5\right)}^{\frac{1}{5}}={\left(2^{\frac{5}{2}}\right)}^{\frac{1}{5}}$  

$x^{5\cdot \frac{1}{5}}=2^{\frac{5}{2}\cdot \frac{1}{5}}$  

$x=2^{\frac{1}{2}}$  

$x=\sqrt{2}$  

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$\text{f)}$  

${\log }_x\left(\frac{1}{12}\right)=-2$  

Z definicji logarytmu:

$x^{-2}=\frac{1}{12}$  

$x^2=12$  

${\left(x^2\right)}^{\frac{1}{2}}={12}^{\frac{1}{2}}$  

$x^{2\cdot \frac{1}{2}}=\sqrt{12}$  

$x=2\sqrt{3}$