Question

Pilne!!!

Proszę o zrobienie zadania z logarytmów :D

Zadanie w załączniku

Answer 1

a) log3 [27 ] = 3 , bo 3^3 = 27

b) log27 [ 3 ] = 1/3, bo 27^(1/3) = 3

c) log3 [ 1/27] = - 3, bo  3^(-3) = 1 / ( 3^3) = 1/27

d) log 1/27 [ 3 ] = - 1/3, bo ( 1/27)^( -1/3) = 27^(1/3) = 3

e) log 1/3 [ 27] = - 3 ,bo (1/3)^(-3) = 3^3 = 27

f) log 27 [ p(3)] = 1/6 , bo 27^(1/6) = p(3)

p(3) - pierwiastek kwadratowy z 3

g) log p(3) [ 27] = 6, bo [ p(3)]^6 = 27

h) log 27 [ 3 p(3)] = x <=> 27^ x = 3 p(3) = 3^(3/2) <=> ( 3^3)^x = 3^(3/2) <=>

<=> 3^(3x) = 3^(3/2) <=> 3x = 3/2  <=> x = 1/2

i) log p(27) [ 3 ] = x <=> [p(3)]^x = 3 <=> [ 3^(1/2)]^x = 3 <=> 3^(0,5 x) = 3 <=>

<=> 0,5 x = 1  <=> x = 2

j)

27^ log 3 [2] = ( 3^3)^log 3 [2] = 3^(3log 3 [2]) = 3^log3[2^3] =

= 3^log3 [ 8] = 8

k)

[p(27)]^log3 [2}] = [ 27^(1/2)]^log3 [2] =[ [ 3^3]^(1/2)]^log3 [2] =

= [ 3^(3/2)]^log3 [2] = 3^ [ (3/2) log 3 [2]] = 3^log 3 [ 2^(3/2)] =

= 2^(3/2) = p(8)

p(27) - pierwiastek kwadratowy z 27

l)

3^( 1 - log3[2]) = 3/ [ 3^log 3[2] ] = 3/2