Fill the missing values to make the equations true.

Answer:

a) [tex]\log_{9} 8 + \log_{9} 11 = \log_{9} 88[/tex]

b) [tex]\log_{7} 11 - \log_{7} 3 = \log_{7} \frac{11}{3}[/tex]

c) [tex]\log_{7}\frac{1}{81} = -4\cdot \log_{7} 3[/tex]

Step-by-step explanation:

Now we proceed to solve each case by appropriate rules for logarithms.

a) The following property shall be used: [tex]\log_{c} a +\log_{c} b = \log_{c} a\cdot b[/tex] ([tex]a = 8[/tex], [tex]b = 11[/tex], [tex]c = 9[/tex])

[tex]\log_{9} 8 + \log_{9} 11 = \log_{9} 8\cdot 11[/tex]

[tex]\log_{9} 8 + \log_{9} 11 = \log_{9} 88[/tex]

b) The following property shall be used: [tex]\log_{c} a - \log_{c} b = \log_{c} \frac{a}{b}[/tex] ([tex]a = 11[/tex], [tex]b = 3[/tex], [tex]c = 7[/tex])

[tex]\log_{7} 11 - \log_{7} 3 = \log_{7} \frac{11}{3}[/tex]

c) The following properties shall be used: [tex]\frac{1}{d} = d^{-1}[/tex], [tex]\log_{c} a^{b}[/tex]

[tex]\log_{7} \frac{1}{81} = \log_{7} 81^{-1}[/tex]

[tex]-\log_{7} 81[/tex]

[tex]-\log_{7} 3^{4}[/tex]

[tex]\log_{7}\frac{1}{81} = -4\cdot \log_{7} 3[/tex]

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