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Domaci zadaci su iz zbirke zadataka iz matematike za ucenike srednjih skola profesora Branislava Bogetica
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Eksponencijalna jednacina
1. 2x=8
2x = 23
X=3
2. 5x=125
5x=53
x=3
3. 3x = 181
3x = 1
34
3x = 3-4
X= - 4
4. 6x = 16x = 60
X= 0
5. (34 ¿x+3= 1
(34 ¿x+3= ( 34 )0
X+3 = 0X= - 3
6. 2-x = 3√42-x = 3√22
2-x = 223
-x =23X= - 23
7. 8x = 5√16(23)x = 5√24
23x = 245
3x = 45
x = 453
=415
8. 16x = 2 (24)x = 24x = 1
X = 14
9. 125x =15
53x = 153x = - 1
X = -1310. (35 )
x = 925
(35 )x = 3
2
52
(35 )x = (35)
2
X = 2
11. ( 436)x = 2168
(22
62)x = 6
3
23
(26)2x = (62)
3
( 26 )2x
= ( 26 )−3
2x = - 3
X = −32
12. (5√121¿x2−8
(5√121¿x2−8 = (5√112)0
X2-8 = 0 a=1, b=0, c=-8x2=8
x1,2=±√8x1,2=±2√2x1=+2√2 x2=−2√2
Domaci: 1155,1156,1157,1159
13. ax-9 = 1ax−9
ax-9 = a-(x-9)
x-9 = -(x-9) x-9 = -x+9 x+x = 9+9 2x = 18 X= 9
14. 3√ax = a3x+22
ax3 = a3x+22
x3 = 3x+22
x∙2 = 3∙(3x+2)
2x = 9x+6
2x-9x = 6
-7x = 6
-x =67 x = -67
15. 100∙102x−1 = 100034
102∙ 102x-1 = (103¿34
102+2x-1 = 1094
2+2x-1 = 94 2x = 94 - 44 2x = 54
X = 54 ∙ 12
X = 58
16. (12)x ∙ (23)
x+1 ∙ (34 )x+2 = 6
(12)x ∙ (23)
x ∙ (23)1 ∙ (34 )x ∙ (34 )2 = 6
( 12 ∙ 23 ∙
34 )x ∙ 23 ∙
916 = 6
(14 )x ∙ 38 = 6
(14 )x = 638
(14 )x = 6 ∙ 83 (14 )x = 16
4-x = 42
-x = 2 X = -2
17. 4√56 ∙ x = 3√5x+2
56− x4 = 5
x+23
6−x4 = x+23 (6-x)∙3 = 4∙(x+2)
18-3x=4x+8 -3x-4x = 8-18 -7x= - 10
X=107
18. 0,125x−0,5
√8 = 8∙(0,25)1-x
¿¿ = 8∙((0,5)2)1-x
¿¿ = 23 ∙ (12 ¿2−2x
¿¿ = 23∙2-(2-2x)
2−3x+1,5
21,5 = 2
3∙2−2+2x
1
2-3x+1,5-1,5 = 23+(-2+2x)
-3x = 3-2+2x -3x-2x = 1 -5x =1
X= -15Domaci: 1166,1167
19. 21∙3x-5x+2 = 9∙3x+2 - 5x+3
5x+3- 5x+2 = 9∙3x+2- 21∙3x
(5x)∙(53)-5x∙52 = 9∙3x∙32-21∙3x
5x∙(125-25) = 3x∙(9∙9-21) 5x∙(100) = 3x∙(60) 5x: 3x = 60:100
(53 ¿x = 35
(53 ¿x = (53 ¿−1
X= - 1
20. 3x+1+ 3x = 108 3x∙(31+1) = 108 3x∙ 4 = 108 3x = 27 3x = 33
Logaritmi
ax =b aeR/{1 } beR+
log2 256 = 8 28 = 256log 100 = log10 100 = 2 102 = 100log b = log10 b
1. Resi logaritme: log7 49 = 2 72 = 49 log2 512 = 9 29 = 512
log 13 81 = log 1
3 34 = log 1
3 ∙(13 ¿4 = - 4
log 14 116 = log 14 142 = log 14 (14 ¿2 = 2 log3 127 = log3 133 = log33-3 = -3 log5 3√5 = log5 513 = 13
2. Izracunati: a) 4 log5 25+2 log3 27 -6 log3 8 b) log3 81+5 log 13 16 – 3 log2 a) 4 log6 25 + 2 log 3 27 -6 log2 8 = 4∙2+2∙3−6 ∙3 = 8+6- 18 = 14-18 = - 4
b) log3 81+5 log 12 16 – 3 log2 132 = log3 ∙ 34 + 5 log 12 ∙ 24 – 3log2 125 = 4+5∙ log 12 ∙ (12 ¿−4- 3 log2 ∙2-5 = 4+5∙(-4)-3∙(-5)= 4-20+15 = -1 3. Odredi X iz jednacina: a) log5 x=0 50=xX=1b) log2 x = -1 2-1 = x X= 12c) log 1
3 x = -2 (13 ¿−2=x X= (31 ¿2= 9 d) logx ∙ 128 = 7 X7 =128 X7 = 27 x= 2e) X3 = 1125 X3 = ¿X= 15
f) X-2 = 116
x-2= 142 x-2 = 4-2x= 4
Domaci: 1186,1187,1188,1189,1190,1191,1192,1193,1195Svojstva Logaritama1.log b = log10 b 2.loga= 03.loga a = 14.loga br = r ∙loga b
5.log as= 1s loga b 6.loga b = 1logb ∙ a7.logb c= log aclogab8.loga= b1∙b2 = loga b1+log b29. loga b1b2 = loga b1- loga b2
Logaritamske funkcijef(x) = loga x a¿0 a≠11.Domen DeR2.Nula funkcije y=0 0=loga x x= a0=1(1,0)
3.presek sa y-osomx=0 y=loga 0 = ay = 0nema resenja...4.tablica i grafik x 1y 0 -1 1 25.znak a¿1 y¿0 xe (1,∞) y¿0 xe (0,1)0¿a<1 y¿0 xe (1,∞) y¿0 xe (0,1)6.monotonosta¿1 y↑ xe R+a¿a¿1 y↓ xeR+7.asimptotaa¿0 y→ - ∞ x→0a¿a¿a y→ ∞ x→0 8.kodomen
DeRf(x)= log 13 x a¿0 a≠11.domenDeR+2.nula funkcije
y=0 0=log x → x= 103=1 (1,0)3.presek sa y-osom x=0 y=log 0 → 1y
3 = 0nema preseka...4.tablica i grafikx 1 3 13 - 1
5y 0 -1 1 2-1=log 1
3 x 5. znak 0<a<1 y<0 xe (1,∞) y>0 xe (-∞,1) 6. 0<a<1 y↓ xeR+ 7. asimptota X=0 0<a<1 y→ ∞ , x→0 8. kodomen DeRf(x)= log2 x 1. domen DeR 2. nula funkcije
y=0 0=log2 x→ x=20=1 (1,0) 3. presek sa y-osom x=0 y=log2 0→ 2y=0 4. tablica i grafik x 1y 0 -1 1 2-1=log2 x 1=log2 x 2=log2 xX= 2-1 = 12 , x=21=2 , x= 22=45. znak a¿1 y¿0 xe (1,∞) y¿<0 xe(-∞,1)6. monotonost a¿1 y↑ xeR+7. asimptota X= 0 a¿0 y→ -∞ x→0
8. kodomenDeRy= log3 x1. domen DeR+2. nule funkcije y= 0 0=log3 x→ x=30=1tacka (1,0)3. presek sa y-osom x=0 y= log3 0 0=3ynema resenja....4. tablica i grafik x 1 1
33 9y 0 -1 1 2
x= 3-1= 13 , x=31=3 , x=32= 95. znak
a¿1 y¿0 xe(1,∞ ¿ y¿0 xe(-∞,1)6. a¿1 y↑ xeR+7. asimptota- je prava y-osa a¿0 y→ -∞ x→08. kodomen DeR
y= log 12 x1. domen DeR2. nula funkcije y= 0 0=log 12 ∙ x → x=(12)0 = 1 tacka (1,0)
3. presek sa y-osom x=0 y= log 12 ∙ 0 , nema resenja...4. tablica i grafik x 1 2 12
14y 0 -1 1 2
5. znak 0¿a<1 y¿0 xe(1,∞ ¿ y¿0 x(-∞ ,1¿6. monotonost 0¿a<1 y↓ xeR 7. asimptota 0¿a<1 y→ ∞ x→ 08. kodomen DeR
Primenom pravila logaritmovanja resiti funkciju: 1. log7 49 = 72 = 22. log2 1 = 03. log2 160 = log2 5∙32 = log2 5∙25 = log2 ∙5 + log2 ∙25 = log5log2 + 5 = 5+0,698970
0,301029 = 5+2,3219038 =7,321935
4. log x100 = log10 x – log10 100 = log x – log10 102 = log x – 25. log 23 8 = log 23 = 16. log 24 4 = log 24 22 = 14 log2 ∙22= 14 ∙ 2 = 127. log7 + 3 log 2 + log 125 – (log 14+log 5) =log7+ log 23 + log 125-(log 14+log5)= log 7∙23-125-(log14
∙5)= log 7 ∙8 ∙125log14 ∙5 = log 100 = log 102 = 28. log 12 4 13 =log 1
2 ∙(22¿13 = log 12 223 = log 12 ∙ (12 ¿
−23 = -23
Logaritmuj primenom pravila:a) * = 3∙(x+y)∙ zb) x4 ∙ y5c) 3√ x87 (6+ y )
a) log2 3 (x+y)∙z = loga 3 + loga ( x+y)+loga z= b) loga x4 ∙ y5 = loga (x4∙y)-loga 5 = (loga x4 + loga y)-loga 5 = (4loga x + loga y )- loga 5 c) loga ∙ 3√ x87 (6+ y ) = loga ( x87 ∙¿ = 13(loga( x87 ∙(6+ y)¿¿= 13(loga x87 + loga (6+y))= 13 ((loga x8 – loga 7) + loga ( 6+y))=13 ((8 loga x – loga 7)+loga (6+y))
Antilogaritmuj:a) 3 log x-2(log y+log z)=3 log x-2(log∙ y ∙ z¿=log x3-log(y∙z)2=log x3
( y ∙ z )2b) 12(log x-3∙(log y -4 log 2))= 12(log x -3∙(log y-log z4))=12 (log x-3∙¿)= 12(logx – log( y
24)3)= 12(log x
¿¿) =
log( x¿¿c) -4log y+9(log(z+1)-log 10)+34 log 5
-4 log y+9∙(log z+110 )+3
4 log 5 = log y-4+log ( z+110 )9+ log(5¿34
=log y-4 ∙ ( z+110 )9 + (5¿34 = log y-4 ∙ ( z+110 )9+ 4√53Domaci: 1205
Logaritamske jednacine1. log3 (3x-5) = 0 uslov: 3x-5>0 3x-5=30 x > 53 3x-5=1 3x=6 X=2
2. log5 (2x-x2) = 0 uslov: 2x-x2 > 0 2x-x2 =1 2x-x2 = 0-x2+2x-1=0 x∙(2−x)= 0X1,2= −b±√b2−4 ac
2a x= 0 2-x =0X1,2= −2±√4−4−(−1 ) ∙(−1)
−2 x=2X1,2= −2±√4−4
−2X1,2= 1 -1 - - -x-0 - + +x-2 - - +F - + - xe(0,2) 3. log5 (x+1)-log5(2x-3) = 1 uslov: x+12x−3>0
x+12x−3 = 1 / ∙2x−3 x+1 > 0
X+1= 2x-3 x> -1 x-2x= -4 2x-3≠0
-x= -4 x≠ 32 X= 4 pripada...
X+1 - + +2x-3 - - +F + - +xe(-∞ ,−1¿U ¿,∞ ¿
4. 1
log x8 +
1log2x 8
+ 1
log4 x8 = 2
log8 x + log8 2x + log8 4x = 2log8 x∙2x ∙4 x=2
log8 8x3 = 2 8x3=82
8x3= 64
X3= 648X3= 23
X=2 Uslov: 8x3>0 X3>0 x>0
5. log3 (x2+1)-log 13 (x2+1) = 4
log3 ( x2+1)-log3−1(x2+1)=4
log3(x2+1)- 1−1log3(x2+1)=4
log3(x2+1)+ log3 (x2+1)=4log3(x2+1) ∙(x2+4)= 4 log3 (x2+1)2 = 4(x2+1)2= 34
(x2+1)2= 81 X2+1 =9 x2+1= -9X2= 8 x2= -10X=± √ 8 x=±√−10
X= ±2√2 nema resenja....
Uslov: (x2+1)2 > 0 xeRKvadratna jednacina uvek ima 2 resenja