Question

Oblicz. (logarytmy)

a) log4 2√8√2
b) log1/2 1/16√2
c) log5³ 25

Załącznik :)

Answer 1

Odpowiedź:

[tex]a)\ \ log_{4}\ 2\sqrt{8\sqrt{2}}=log_{4}\ 2\sqrt{2^3\cdot2^{\frac{1}{2}}}=log_{4}\ 2\sqrt{2^{3+\frac{1}{2}}}=log_{4}\ 2\sqrt{2^{\frac{7}{2}}}=\\\\=log_{4}\ 2\cdot(2^{\frac{7}{2}})^\frac{1}{2}}=log_{4}\ 2\cdot2^{\frac{7}{4}}=log_{4}\ 2^1\cdot2^{\frac{7}{4}}=log_{4}\ 2^{1+\frac{7}{4}}=log_{2^2}\ 2^{\frac{11}{4}}=\\\\=\dfrac{\frac{11}{4}}{2}\cdot log_{2}2=\frac{11}{4}:2\cdot1=\frac{11}{4}\cdot\frac{1}{2}=\frac{11}{8}=1\frac{3}{8}[/tex]

[tex]b)\ \ log_{\frac{1}{2}}}\ \frac{1}{16\sqrt{2}}=log_{2^{-1}}\ (16\sqrt{2})^{-1}=log_{2^{-1}}\ (2^4\cdot2^{\frac{1}{2}})^{-1}=log_{2^{-1}}\ (2^{4+\frac{1}{2}})^{-1}=\\\\=log_{2^{-1}}\ (2^{\frac{9}{2}})^{-1}=log_{2^{-1}}\ 2^{-\frac{9}{2}}=\dfrac{-\frac{9}{2}}{-1}\cdot log_{2}2=\frac{9}{2}\cdot1=\frac{9}{2}=4\frac{1}{2}[/tex]

[tex]c)\ \ log_{5^3}\ 25=log_{5^3}\ 5^2=\frac{2}{3}\cdot log_{5}5=\frac{2}{3}\cdot1=\frac{2}{3}[/tex]

Szczegółowe wyjaśnienie:

Zastosowałam wzory

[tex]\sqrt[n]{a^m}=a^{\frac{m}{n}}\\\\a^m\cdot a^n=a^{m+n}\\\\(\frac{1}{a})^n=a^{-n}\\\\log_{a^y}\ b^x=\frac{x}{y}\cdot log_{a}b[/tex]