Odpowiedź:
e ) = log[tex]_{1/5}[/tex] [tex]\frac{2}{\frac{2}{25} } = log_{1/5} 25 = - 2[/tex]
bo ( 1/5 ) [tex]^{-2} = 5^2 = 25[/tex]
f ) = log[tex]_3[/tex] ( 5*[tex]\frac{27}{5} ) =[/tex] [tex]log_3 27 = 3[/tex] bo 3³ = 27
g ) = [tex]log_{1/5} \frac{7}{\frac{7}{5} } = log_{1/5} 5 = - 1[/tex] bo ( [tex]\frac{1}{5} )^{-1} = 5[/tex]
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Verified answer
Odpowiedź:
e ) = log[tex]_{1/5}[/tex] [tex]\frac{2}{\frac{2}{25} } = log_{1/5} 25 = - 2[/tex]
bo ( 1/5 ) [tex]^{-2} = 5^2 = 25[/tex]
f ) = log[tex]_3[/tex] ( 5*[tex]\frac{27}{5} ) =[/tex] [tex]log_3 27 = 3[/tex] bo 3³ = 27
g ) = [tex]log_{1/5} \frac{7}{\frac{7}{5} } = log_{1/5} 5 = - 1[/tex] bo ( [tex]\frac{1}{5} )^{-1} = 5[/tex]
Szczegółowe wyjaśnienie: