Giải phương trình
3
alog2x — đã
2
x2rlog2 —8
sau: 2” = 33/21
log E1) =lgl,5
32x+2 -L22y +2 —17
23x+l11327—=8
loga(2x2).log22 =]
log x = log, («+2)
gl 2% 1 4 19829 —6
lgx_ zxn¬_ „lg5
fogz9—53 x
3-x `
log1—2x
g 83 “ —n x24
lgx+7
log2x2— 4x43 =2
„hin/z| —
5—12
log et 1085 12x =
x tex — 10x
log ox = log,(v2x+1 —])
logx+, flog2x+1—5 =0
.1fõ+x + VIT-2x=2
log„(x2+3x+2)+log„(x2+7x+12) = 3+loga3
log, («-Vx2—-1).log.(«+V¥x2-1) =
log, _»,,(6x?—-5x+1)-log, 4 (4x?-4x+1)—2 =0
2x2+x_— 4,2X2—~x— 22x +44 = 0, 82 x+log,x+log 4X = logy ox
(v3-v2)”+(vVä+v2)” =(v5)”
4ax+12x+4_— 2x+2-+16
2x22x—Ä3_— ax—Ìl
2*.3x—lsX—2—=12
x2—6x—3 ——
(2) —3x—l2_—1 |
(25)*-1=4
2*.3ax+1=(v3)X+2
x
10.2⁄—~1_22x—l_6.22<0
4K = y AR
9x—1_9x2-x — (x-1)
log„x = log„(vx +2)
—1
5% 8° =500 (*)
logg2tlog, (4x) =3
9x—1_9x2-x — (x-1)
314 3 — đÌo x2Ìog„x
4'95x 897% — “1°83
log, xtlog,x = 1+log,x.log,x aled0x)_ elgx — 2 qlg(00x2)
Ig4(x—1)2+Ig2(œ—1)° =25
Qx—1_9x?—x = (x-1)?
7 [81S] _ |cosx|
3
glog 3 — 3Ìoga„x = 2Ìogax
log x+log.x = (log 3).(log 225)
logaa@—v x2~1)
3log„(1+v#+Äz) = 2log„v%
x 1 12
23x _6.2 “Be j†2x=1,
loại ogx— —2+9”]— 2x
logi8~2)— 1ole(3—>)
O41) 80H) 100(-L1)
(8+3/7)!#X-+(@—3V7)*6* =16 (*) (v1—x +v1+x)log„(x2->) =0 log2(3* —1)-4] og„(3” —1)4+3=0
aled0x)_ elgx — 2 qlg(00x2) logax+log„x = 1+logax.logax Ig4(x—1)2+Ig2(œ—1)° = 25
el2*—5|_elx—1| — L_ _1_
~ |2x—5| |x—-l|
]
log (3x-) toga ya)? — 2+log,(x+1)
3)
1 83x_ 25
3+ipg—x =log, Í 2 -25 )
x 1 12
29x _6.2 Seay te =!
loga*+15.2°+27 +2logzz-ã
2
Trang 23
slog, (x+2)?+3 = log, (4-x)* +log,(x+6)°
2(1og4x)° =[log,x]|log,(V2%F1-1}
Ax2— 3x+2-Lax2+6x-+Š — 42x2+3x+7 44
(2+v3)”+(7+4v3)(2- V3)" =42-+V3) (*)
Gc—4)"log , (<—1)—2log 4 (x 1) = (x- 4) log —14—2log „¡16
g2x—1432x452x+1_ 9 43xtlis5xt2
3 logi (+2)? — 3= logs (4x)? + log1 (x+6)°
(v3—-V2)" + (V3-+V2)" = (vB)
loga*+15.2°+27 +2logz.z=3
loge(x2—5x+6)” = slog /sŠgˆ +logalx~3|
2Inx+In(2x-3) =0
cCOẽx
loga(3.2”—1)=2x+1
+loga(9”~—6) = log„(4.3”—6)
log[V2(x3+x2)~2]+log„(2x+2) =0
3
loga{ 2log,{1+log,(1+3log,x)]} = 5
log, x — logi a + 9 log _ 4 logi x
log„(x2+3x+2)+log„(x2+7x+12) =3+log„3
2
log2
x 3" —Blog.x+7 = =
¥x+1-1 ~ ¥x+i4+1 aa
(2 =1+x2
log (3x1) + ioe gaya = 2 tS get) 5 logi (x42)? _ = log1 (4—x)9 + logi (x+6)°
42x2_9 4x?+x442x—Q log2 2+log,4x=3 log rpg? S 108 yz?
("47 @) Fs <0
lg(3x—-y) +lg@+y) —41g2 = 0
Trang 3log„(x2 +y2)-log„2x+1 = log, («+3y)
log, (xyt+1)—log, @y?2+2y—2x+4) = loguÿ-1