Question

Proszę o rozwiązanie daje NAJ
najlepiej z obliczeniami

Answer 1

Odpowiedź:

2.45

[tex]2[/tex] [tex]log_6 2 = log_6 2^2 = log_6 4[/tex]

oraz

2 - [tex]log_6 9 = log_6 36 - log_6 9 = log_6 \frac{36}{9} = log_6 4[/tex]

Odp. A

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2.46

[tex]a_n = 2018 + log_3( n + 1)\\więc\\a_{23} - a_7 = 2018 + log_3 24 - ( 2018 + log_3 8 ) = log_3 24 - log_3 8 = log_3 \frac{24}{8} = log_3 3 = 1[/tex]

Odp. D

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2.47

[tex]a_n = log_2 ( 10 +2 n)[/tex]

[tex]a_{11} = log_2 ( 10 + 2*11) = log_2 32 = 5[/tex]

Odp. C

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2.48

[tex]a_n = log_4 ( 11 n - 35 )[/tex]    n  ≥ 4

więc

[tex]a_9 = log_4 ( 11*9 - 35) = log_4 64 = 3[/tex]

Odp. A

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2.49

n  ≥ 1            [tex]a_n = 6 + log_2 ( 24 n - 16 )[/tex]

[tex]a_1 = 6 + log_2 ( 24*1 - 16) = 6 + log_2 8 = 6 + 3 = 9[/tex]

[tex]a_2 = 6 + log_2 ( 24*2 - 16 ) = 6 + log_2 32 = 6 + 5 = 11[/tex]

[tex]a_1 + a_ 2 = 9 + 11 = 20[/tex]

Odp.  B

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2.50

[tex]a_n = log ( \frac{1}{n} )[/tex]                 n ≥ 1

więc

[tex]a_{100} = log ( \frac{1}{100} ) = - 2[/tex]      bo   [tex]10^{-2} = \frac{1}{10^2} = \frac{1}{100}[/tex]

Odp. A        bo  - 2  <  - 1

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Szczegółowe wyjaśnienie: