[tex]c=log_{2} 9=log_{2}3^{2} =2\cdot log_{2}3\\\\\dfrac{1}{2} c=log_{2}3\\\\a)\\\\log_{3} 4=\dfrac{log_{2}4 }{log_{2} 3} =\dfrac{log_{2}2^{2} }{log_{2} 3} =\dfrac{2\cdot log_{2}2 }{log_{2} 3} =\dfrac{2\cdot 1 }{log_{2} 3} =\dfrac{2 }{log_{2} 3}\\\\\\log_{3} 4=\dfrac{2 }{log_{2} 3}~~\land~~\dfrac{1}{2} c=log_{2}3~~\Rightarrow ~~log_{3} 4=\dfrac{2}{\dfrac{c}{2} } =2\div \dfrac{c}{2} =2\cdot \dfrac{2}{c} =\dfrac{4}{c}=4\cdot c^{-1} ~~zal.~~c\neq 0[/tex]
[tex]b)\\\\log_{\frac{1}{9} } \dfrac{1}{8} =\dfrac{log_{2} \frac{1}{8} }{log_{2 }\frac{1}{9} } =\dfrac{log_{2} 8^{-1} }{log_{2 }9^{-1} } =\dfrac{log_{2} 2^{-3} }{-1\cdot log_{2 }9 } =\dfrac{-3\cdot log_{2} 2 }{-1\cdot log_{2 }9 } =\dfrac{-3\cdot 1 }{-1\cdot log_{2 }9 }=\dfrac{-3 }{-1\cdot log_{2 }9 }=\dfrac{3 }{log_{2 }9 }\\\\\\log_{\frac{1}{9} } \dfrac{1}{8} =\dfrac{3 }{log_{2 }9 }~~\land~~log_{2} 9=c~~\Rightarrow ~~log_{\frac{1}{9} } \dfrac{1}{8} =\dfrac{3}{c}=3\cdot c^{-1} ~~zal.~~c\neq 0[/tex]
korzystałam ze wzorów:
[tex]log_{x} x=1\\\\log_{a} b=\dfrac{log_{c} b}{log_{c} a} \\\\log_{a} x^{n} =n\cdot log_{a} x[/tex]
