Question

1. Sprawdz czy poprawnie obliczono logarytmy

log6 = 1

log0,001= -3

log12 1 = 0

log27 3 = 1/3

log2 pierwiastek z 2 = 1/2

log81 27= 3/4

2. Wyznacz x. Podaj odpowiednie założenia

A: log2x = 3

log32x = 2/5

log16x = 2

logx = 2/3

logx = -0,5

B:logx 27 = 3

loggx (1/8) = -3

logx 16 = 4 l

ogx3 = 1/3

logx81 = 4/3

C: log2 8 pierwiastków z 2 = x

logx x do potegi 5 = 5

logx x do potęgi -3 = -3

logx x do potegi 2/3 = 2/3

Bardzo proszę o pomoc wgl tego nie rozumiem :(

Answer 1

z.1

log 6 = 1  

  żle, bo 10^1 = 10

log 0,001 = - 3 

dobrze, bo 10^(-3) = 1/( 10^3) = 1 / 1000 = 0,001

log12 [ 1] = 0 

dobrze, bo 12^0 = 1

log 27[ 3] = 1/3 

dobrze, bo    27^(1/3) = 3, 3^3 = 27

log 2 [ p(2)] = 1/2

dobrze, bo 2^(1/2) = p(2)

p(2) - pierwiastek kwadratowy z 2

log 81 [ 27] = 3/4

dobrze

bo  81^(3/4) = [ 81^(1/4)]^3 = 3^3 = 27

=========================================

z.2

a)  log 2 [ x] = 3   ;  x > 0

x = 2^3 = 8

============

log32 [ x] = 2/5 ; x > 0

x = 32^(2/5) = [ 32^(1/5)]^2 = 2^2 = 4

======================================

log 16 [ x] = 2 ;  x > 0

x = 16^2 = 256

================

log x = 2/3 ;  x > 0

x = 10^(2/3)

============

log x = - 1/2;   x > 0

x = 10^(-1/2) = 1/[ 10^(1/2)] = 1 / p(10)

 

p(10) - pierwiastek kwadratowy z 10

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b)

 

log x [27 ] = 3

x^3 = 27 =3^3

x = 3

=====

log x ( 1/8) = - 3

x^(-3) = 1/8 = (1/2)^3 = 2^(-3)

x = 2

=====

log x [ 3 ] = 1/3

x^(1/3) = 3

x = 3^3 = 27

===============

log x [81] =  4/3

x^(4/3) = 81

x^(1/3) = 3

x = 3^3

x = 27

==============

c)

log 2 [ 8 p(2)] = x

2^x = 8 p(2) = p(2)^7 = [ 2^(1/2)]^7 = 2^(7/2)

x = 7/2 = 3,5

============

log x[ x^5] = 5

5* log x [ x] = 5

5*1 = 5  m

x  > 0  i  x  róne od  1

=========================

log x [ x^(-3)] = - 3

x^(-3) = x^(-3)

x > 0  i  x  różne od 1

=====================

log x [ x^(2/3) ] = 2/3

x^(2/3) = x^(2/3)

x > 0 i  x  różne od  1

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