Question

PILNIE NA DZIS POMOZCIE PLIS

Answer 1

z.1

log 2/3 [ 4 ] - 2 log 2/3 [ 3 ] = log 2/3 [ 4 ] - log 2/3  [ 3^2] =

= log 2/3 [ 4 ] - log 2/3  [ 9 ] =  log 2/3  [ 4 / 9] = 2  , bo ( 2/3)^2 = 4/9

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log 5 [ 16 ] - log 5 [ 80] = log 5 [ 16/80] = log 5 [ 1/5] = - 1 , bo  5^(-1) = 1/5

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log 4 [ 16] + log 4 [ 64 ] = 2 + 3 = 5  , bo  4^2 = 16  i  4^3 = 64

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z.3

log 3 [ x^2 + 5] = 2

log 3 [ x^2 + 5] = log 3 [9 ]

x^2 + 5 = 9

x^2 = 9 - 5 = 4

x = - 2   lub   x = 2

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z.4

A = ( 1; 2),  B = ( 1/9 ; 0) ,  C = ( 3; 3)

y = log 3 [ x ] + 2

zatem

Dla x = 1 mamy  y = 3 [ 1] + 2  = 0 + 2 = 2

A  należy do wykresu tej funkcji.

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Dla x = 1/9  mamy   y = log 3 [ 1/9] + 2 = - 2 + 2 = 0

B  należy do wykresu danej funkcji.

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Dla x = 3  mamy  y = log 3 [ 3 ] + 2 = 1 + 2 = 3

C  należy do wykresu tej funkcji.

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log 3 [ 1 ] = 0 , bo   3^0 = 1

log 3 [ 1/9] = - 2 , bo  3^(-2) = 1 / ( 3^2) = 1/9

log 3 [ 3] = 1 , bp 3^1 = 3

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