Question

Sprawdzian z logarytmów, wszystkie zadania.
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Answer 1

Odpowiedź:

1.

a)

[tex]log_2 448 + log_2 12 - log_2 168 = log_2 \frac{448*12}{168} = log_2 32 = 5[/tex]                bo [tex]2^5 = 32[/tex]

b )

[tex]log_{1/5} 25^{-3} - log_3 27^{5/2} - log_}1/3} ( log 1000^9) =\\= - 3* log_{1/5} 25 - \frac{5}{2} log_327 - log_{1/3} ( 9*3) =\\= - 3*( - 2) - \frac{5}{2}*3 - log_{1/3} 27 = 6 - 7,5 - ( - 3) = 9 - 7,5 = 1,5[/tex]

c )  [tex]log_{1/3} 162 + log_{1/3} 21 - 3 log_{1/3} 84 = log_{1/3} 162^6 + log_{1/3} 21^3 - log_{1/3} 84^3 =[/tex]

[tex]= log_{1/3} \frac{162^6*21^3}{84^3} = log_{1/3} \frac{162^6}{64}[/tex] [tex]= 3* log_{3^{-1}} \frac{162^2}{4} = -3 log_3 6561[/tex]

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3.

f ( x) = [tex]log_{1/16} x[/tex]                    <  [tex]\frac{1}{2} ; \sqrt{2} >[/tex]

Funkcja  jest malejąca, więc

[tex]y_{max} = f ( 1/2) = log_{1/16} \frac{1}{2} = \frac{1}{4}[/tex]          bo  [tex]( \frac{1}{16} )^{1/4} = \sqrt[4]{\frac{1}{16} } = \frac{1}{2}[/tex]

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[tex]y_{min} = f (\sqrt{2} ) = log_{1/16} \sqrt{2} = log_{1/16} 2^{1/2} = \frac{1}{2} *( - \frac{1}{4} ) = - \frac{1}{8}[/tex]

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4.

f ( x) = [tex]log_{1/2} ( x + 1 ) - 5[/tex]

x + 1 > 0  ⇒  x >  - 1

Df = ( - 1; + ∞ )

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Asymptota  pionowa :  x =  - 1

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Funkcja  jest  malejąca.

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f ( - 1/2) = [tex]log_{1/2} \frac{1}{2} - 5 = 1 - 5 = - 4[/tex]        A = ( -1/2 , - 4 )

f ( 0 ) = [tex]log_{1/2} 1 - 5 = 0 - 5 = - 5[/tex]            B = (  0, - 5 )

f (1 ) = [tex]log_{1/2} 2 - 5 = - 1 - 5 = -6[/tex]           C = ( 1, - 6 )

f ( 3) = [tex]log_{1/2} 4 - 5 = - 2 - 5 = - 7[/tex]          D = ( 3 , - 7 )

Narysuj wykres -   linia przechodząca przez punkty A, B, C, D

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5.

a )   [tex]3^{- 3- 0,4}* 9^{3 + 0,2} = 3^{- 3 - 0,4}* (3^2)^{ 3 + 0,2} = 3^{-3,4}*3^{6,4} = 3^3 = 27[/tex]

b )  [tex]( \frac{1}{3} )^{-5} : 27^{0,3} : \frac{1}{3^{10}} = 3^5 : ( 3^3)^{0,3 }: 3^{-10} = 3^5 : 3^{0,9} : 3^{-10} =[/tex]

      [tex]= 3^{5 - 0,9} : 3^{-10} = 3^{4,1} : 3^{- 10} = 3^{4,1 - ( - 10)} = 3^{14,1}[/tex]

c )  [tex][ ( 81^{-3/8})^{4/3}]^{-2} = [ 81^{-3/8*4/3}]^{-2} = [ 81^{-1/2}]^{-2} = 81^{1} = 81[/tex]

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6.

[tex]log_{16} x = \frac{log_3 1/27 + log_3 ( 3\sqrt{3} )}{log_2 16 - log_2 8} = \frac{- 3 +1.5}{log_2 ( 16 : 8 )}[/tex]   [tex]= \frac{-1,5}{log_2 2} = - 1,5[/tex]

x = [tex]16^{-1,5} = \frac{1}{16^{1,5}} = \frac{1}{16*4} = \frac{1}{64}[/tex]

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7.

a )     [tex]f ( x) = log_9 x[/tex]    P = ( [tex]\frac{1}{\sqrt{3} } ,[/tex] t )

więc   [tex]log_9 \frac{1}{\sqrt{3} } = log_{3^2} 3^{-1/2} = \frac{1}{2} *(-\frac{1}{2} )*log_3 3= - \frac{1}{4} *1 = - \frac{1}{4}[/tex]

t = - [tex]\frac{1}{4}[/tex]

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b )   [tex]f ( x) = log_{1/2} x[/tex]          P = ( [tex]\frac{1}{t}[/tex] , - 3 )

więc

[tex]log_{1/2} \frac{1}{t} = - 3[/tex]   ⇒   [tex]\frac{1}{t} = ( 1/ 2)^{-3} = 2^3 = 8\\t = \frac{1}{8}[/tex]

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8.

P = ( - 1 [tex]\frac{1}{3}[/tex] ,  p )                    Q = ( q ,  125 )

f ( x) =  ( 5 √5 [tex])^x[/tex]

więc

p  = ( 5 [tex]\sqrt{5} )^{-4/3} = ( 5 ^{3/2})^{-4/3} = 5^{- 2} = \frac{1}{5^2} = \frac{1}{25}[/tex]

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oraz

[tex]( 5[/tex] [tex]\sqrt{5} )^q = 125 = 5^3[/tex]

[tex](5^{1,5})^q = 5^3\\1,5 q = 3\\q = 2[/tex]

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