bardzo proszę o pomoc, nie mogę sobie poradzić z rozgryzieniem tych zadań.przykłady w załączniku.
z.1
2 log x + log ( 6 - x^2 ) = 0
Założenia:
x > 0 i 6 - x^2 > 0
x > 0 i 6 > x^2
x > 0 i x^2 < 6
x > 0 i x < p(6)
zatem
log x^2 + log ( 6 - x^2) = 0
log [ x^2 *(6 - x^2)] = log 1
x^2 * ( 6 - x^2 ) = 1
6 x^2 - x^4 - 1 = 0
x^4 - 6 x^2 + 1 = 0
-----------------------
Podstawienie y =x^2
Mamy
y^2 - 6 y + 1 = 0
delta = (-6)^2 - 4*1*1 = 36 - 4 = 32 = 16*2
p (delty ) = 4 p(2)
y = [ 6 - 4 p(2)]/2 = 3 - 2 p(2)
lub
y = [ 6 + 4 p(2)]/2 = 3 + 3 p(2)
czyli
x^2 = 3 - 2 p(2) > 0 lub x^2 = 3 + 2 p(2) > 0
x = p [ 3 - 2 p(2)] lub x = p [ 3 + 2 p(2) ]
===========================================
Ponieważ 3 - 2 p(2) > 0 , więc p [ 3 - 2 p(2) ] > 0
oraz 3 + 2 p(2) < 6 , więc p [ 3 + 2 p (2) ] < p(6)
==================================================
z.2
[ 2 log x ] / [ log ( 5x - 4 ) ] = 1
Założenia: x > 0 i 5x - 4 > 0
czyli x > 0 i 5 x > 4
x > 0 i x > 4/5 = 0,8
x > 0,8
------------
Mamy więc
2 log x = log ( 5 x - 4)
log x^2 = log ( 5 x - 4)
x^2 = 5x - 4
x^2 - 5 x + 4 = 0
----------------------
delta = (-5)^2 - 4*1*4 = 25 - 16 = 9
p( delty ) = p ( 9) = 3
x1 = [ 5 - 3 ]/2 = 2/2 = 1 > 0,8
x2 = [ 5 + 3 ]/2 = 8/2 = 4 > 0,8
Odp. x1 = 1; x2 = 4
====================
z.3
log ( 3x - 91 ) - log ( 30 - x) = 1
3x - 91 > 0 i 30 - x > 0
3x > 91 i 30 > x
x > 30 1/3 i x < 30 - sprzeczność
Równanie nie ma rozwiązania.
==============================
z.4
3 ^ log x = 1/27
3^log x = 3 ^(-3)
log x = - 3
x = 10^(-3) = 1/10^3 = 1/1000 = 0,001
===================================
z.5
log [ 2x / ( 4x - 15 )] = 2
założenie:
2x /( 4x - 15 ) > 0
2x > 0 i 4x - 15 > 0 lub 2x < 0 i 4x - 15 < 0
x > 0 i x > 15/4 lub x < 0 i x < 15/4
x > 15/4 lub x < 0
--------------------------------
log [ 2x / ( 4x - 15 ) ] = log 100
2x/( 4x - 15) = 100
2x = 100*(4x - 15)
2x = 400 x - 1500
398 x = 1 500
x = 1 500 / 398 = 750 / 199 = około 39,47 > 15/4
==============
Odp. x = 750/199
z.6
log 4 { 2 log 3 [ 1 + log 2 ( 1 + log 2 ( x))] } = 1/2
2 log 3 [ 1 + log 2 ( 1 + log 2(x))] = 4^(1/2) = 2 / : 2
log 3 [ 1 + log 2 ( 1 + log 2 (x))] = 1
1 + log 2 ( 1 + log 2 (x)) = 3
log 2 ( 1 + log 2 (x)) = 2
1 + log 2(x) = 2^2 = 4
log 2 (x) = 4 -1 = 3
x = 2^3 = 8
=================
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z.1
2 log x + log ( 6 - x^2 ) = 0
Założenia:
x > 0 i 6 - x^2 > 0
x > 0 i 6 > x^2
x > 0 i x^2 < 6
x > 0 i x < p(6)
zatem
log x^2 + log ( 6 - x^2) = 0
log [ x^2 *(6 - x^2)] = log 1
x^2 * ( 6 - x^2 ) = 1
6 x^2 - x^4 - 1 = 0
x^4 - 6 x^2 + 1 = 0
-----------------------
Podstawienie y =x^2
Mamy
y^2 - 6 y + 1 = 0
delta = (-6)^2 - 4*1*1 = 36 - 4 = 32 = 16*2
p (delty ) = 4 p(2)
y = [ 6 - 4 p(2)]/2 = 3 - 2 p(2)
lub
y = [ 6 + 4 p(2)]/2 = 3 + 3 p(2)
czyli
x^2 = 3 - 2 p(2) > 0 lub x^2 = 3 + 2 p(2) > 0
x = p [ 3 - 2 p(2)] lub x = p [ 3 + 2 p(2) ]
===========================================
Ponieważ 3 - 2 p(2) > 0 , więc p [ 3 - 2 p(2) ] > 0
oraz 3 + 2 p(2) < 6 , więc p [ 3 + 2 p (2) ] < p(6)
==================================================
z.2
[ 2 log x ] / [ log ( 5x - 4 ) ] = 1
Założenia: x > 0 i 5x - 4 > 0
czyli x > 0 i 5 x > 4
x > 0 i x > 4/5 = 0,8
x > 0,8
------------
Mamy więc
2 log x = log ( 5 x - 4)
log x^2 = log ( 5 x - 4)
x^2 = 5x - 4
x^2 - 5 x + 4 = 0
----------------------
delta = (-5)^2 - 4*1*4 = 25 - 16 = 9
p( delty ) = p ( 9) = 3
x1 = [ 5 - 3 ]/2 = 2/2 = 1 > 0,8
x2 = [ 5 + 3 ]/2 = 8/2 = 4 > 0,8
Odp. x1 = 1; x2 = 4
====================
z.3
log ( 3x - 91 ) - log ( 30 - x) = 1
Założenia:
3x - 91 > 0 i 30 - x > 0
3x > 91 i 30 > x
x > 30 1/3 i x < 30 - sprzeczność
Równanie nie ma rozwiązania.
==============================
z.4
3 ^ log x = 1/27
3^log x = 3 ^(-3)
log x = - 3
x = 10^(-3) = 1/10^3 = 1/1000 = 0,001
===================================
z.5
log [ 2x / ( 4x - 15 )] = 2
założenie:
2x /( 4x - 15 ) > 0
2x > 0 i 4x - 15 > 0 lub 2x < 0 i 4x - 15 < 0
x > 0 i x > 15/4 lub x < 0 i x < 15/4
x > 15/4 lub x < 0
--------------------------------
zatem
log [ 2x / ( 4x - 15 ) ] = log 100
2x/( 4x - 15) = 100
2x = 100*(4x - 15)
2x = 400 x - 1500
398 x = 1 500
x = 1 500 / 398 = 750 / 199 = około 39,47 > 15/4
==============
Odp. x = 750/199
====================
z.6
log 4 { 2 log 3 [ 1 + log 2 ( 1 + log 2 ( x))] } = 1/2
2 log 3 [ 1 + log 2 ( 1 + log 2(x))] = 4^(1/2) = 2 / : 2
log 3 [ 1 + log 2 ( 1 + log 2 (x))] = 1
1 + log 2 ( 1 + log 2 (x)) = 3
log 2 ( 1 + log 2 (x)) = 2
1 + log 2(x) = 2^2 = 4
log 2 (x) = 4 -1 = 3
x = 2^3 = 8
=================