Odpowiedź:
z.18
[tex]f ( x) = log_ a ( x - 2 ) + 1[/tex]
a) P ( 4, 2)
więc
[tex]log_a ( 4 - 2) + 1 = 2[/tex]
[tex]log_a 2 = 1[/tex]
a = 2 bo [tex]2^1 = 2[/tex]
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b ) P ( 6, 2)
[tex]log_a (6 - 2) + 1 = 2[/tex]
[tex]log_a 4 = 1[/tex]
a = 4 bo [tex]4^1 = 4[/tex]
c ) P ( 7/3 , 0 )
[tex]log_a ( \frac{7}{3} - 2) + 1 = 0[/tex]
[tex]log_a \frac{1}{3} = - 1[/tex]
a = 3 bo [tex]3^{-1} = \frac{1}{3}[/tex]
d ) P( 4, 3 )
[tex]log_a ( 4 - 2) + 1 = 3[/tex]
[tex]log_a 2 = 2[/tex]
a = [tex]\sqrt{2}[/tex] bo ( √2 )² = 2
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Szczegółowe wyjaśnienie:
rozwiązanie w załączniku
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Verified answer
Odpowiedź:
z.18
[tex]f ( x) = log_ a ( x - 2 ) + 1[/tex]
a) P ( 4, 2)
więc
[tex]log_a ( 4 - 2) + 1 = 2[/tex]
[tex]log_a 2 = 1[/tex]
a = 2 bo [tex]2^1 = 2[/tex]
=====
b ) P ( 6, 2)
[tex]log_a (6 - 2) + 1 = 2[/tex]
[tex]log_a 4 = 1[/tex]
a = 4 bo [tex]4^1 = 4[/tex]
=====
c ) P ( 7/3 , 0 )
[tex]log_a ( \frac{7}{3} - 2) + 1 = 0[/tex]
[tex]log_a \frac{1}{3} = - 1[/tex]
a = 3 bo [tex]3^{-1} = \frac{1}{3}[/tex]
=====
d ) P( 4, 3 )
[tex]log_a ( 4 - 2) + 1 = 3[/tex]
[tex]log_a 2 = 2[/tex]
a = [tex]\sqrt{2}[/tex] bo ( √2 )² = 2
======
Szczegółowe wyjaśnienie:
Odpowiedź:
rozwiązanie w załączniku