Korzystamy z definicji logarytmu, podstawowych własności oraz praw działań na logarytmach i potęgach (patrz zał. 1 i 2).
Zad. 1
a)
[tex]log_232 =x \\ 2^x = 32 \\ 2^x= 2^5 \\ x = 5 \\ log_232 =5[/tex]
b)
[tex]log0,01 =x \\ 10^x = 0,01 \\ 10^x= \frac{1}{100}\\ 10^x= \frac{1}{10^2} \\ 10^x= 10^{-2} \\ x = -2 \\ log0,01 =-2[/tex]
c)
[tex]log_3 1=0[/tex]
d)
[tex]log_6 \frac{1}{216} =x \\ 6^x= \frac{1}{216}\\ 6^x= \frac{1}{6^3} \\ 6^x= 6^{-3} \\ x = -3 \\ log_6 \frac{1}{216} =-3[/tex]
e)
[tex]log_{\sqrt{3}} 9=x \\ (\sqrt{3})^x=9 \\ (3^{\frac{1}{2}})^x=3^2 \\ 3^{\frac{1}{2}x}=3^2 \\ \frac{1}{2}x = 2 \ \ \ |\cdot 2 \\ x = 4 \\ log_{\sqrt{3}} 9=4[/tex]
f)
[tex]log_{\frac{1}{3}} 81=x \\ (\frac{1}{3})^x=81 \\ (3^{-1})^x= 3^4 \\ 3^{-x}= 3^4 \\ -x = 4 \ \ \ |\cdot (-1) \\ x = - 4 \\ log_{\frac{1}{3}} 81= -4[/tex]
g)
[tex]log_{0,25} \sqrt{2} =x \\ (0,25)^x=\sqrt{2} \\ (\frac{1}{4})^x=2^{\frac{1}{2}} \\ (\frac{1}{4})^x=(\sqrt{4})^{\frac{1}{2}} \\ (\frac{1}{4})^x=(4^{\frac{1}{2}})^{\frac{1}{2}} \\ (\frac{1}{4})^x=4^{\frac{1}{4}} \\ (\frac{1}{4})^x=((\frac{1}{4})^{-1})^{\frac{1}{4}} \\ (\frac{1}{4})^x=(\frac{1}{4})^{-\frac{1}{4}} \\ x = -\frac{1}{4} \\ log_{0,25} \sqrt{2} = -\frac{1}{4}[/tex]
h)
[tex]log_{0,2} \sqrt[3]{5} =x \\ (0,2)^x= \sqrt[3]{5} \\ (\frac{2}{10})^x= 5^{\frac{1}{3}} \\ (\frac{1}{5})^x= ((\frac{1}{5})^{-1})^{\frac{1}{3}} \\ (\frac{1}{5})^x= (\frac{1}{5})^{-\frac{1}{3}} \\ x =-\frac{1}{3} \\ log_{0,2} \sqrt[3]{5} =-\frac{1}{3}[/tex]
Zad. 2
a)
[tex]log_33^{20}=20 \cdot log_33 = 20 \cdot 1 = 20[/tex]
b)
[tex]log_55^{0,1}=0,1 \cdot log_55 = 0,1 \cdot 1 = 0,1[/tex]
c)
[tex]log_{0,3}\sqrt{0,3}=log_{0,3} (0,3)^{\frac{1}{2}}=\frac{1}{2}\cdot log_{0,3}0,3 = \frac{1}{2}\cdot 1 =\frac{1}{2}[/tex]
d)
[tex]log_{\sqrt{3}}\sqrt{3^{10}} =log_{\sqrt{3}}(\sqrt{3})^{10}= 10 \cdot log_{\sqrt{3}}\sqrt{3}= 10 \cdot 1 = 10[/tex]
e)
[tex]4^{log_45}=5[/tex]
f)
[tex]3^{7log_32}=3^{log_32^7}=2^7 = 128[/tex]
g)
[tex]25^{log_53}=(5^2)^{log_53}=5^{2 \cdot log_53}=5^{log_53^2}=3^2=9[/tex]
h)
[tex]7^{log_{49}4}=(\sqrt{49})^{log_{49}4}=(49^{\frac{1}{2}})^{log_{49}4}=49^{\frac{1}{2} \cdot log_{49}4}= 49^{log_{49}4^{\frac{1}{2}}}=49^{log_{49}\sqrt{4}}=\\ = 49^{log_{49}2}=2[/tex]
