Logaritmos
58
Profesor: Carlos Mario Calle
-
Upload
carlos-calle -
Category
Education
-
view
274 -
download
0
Transcript of Logaritmos
Profesor: Carlos Mario Calle
Caso I
log3 (2 x - 1) = 2
32 = 2 x - 1log3 (2 x - 1) = 2
32 = 2 x - 1
9 = 2 x - 1
log3 (2 x - 1) = 2
32 = 2 x - 1
9 = 2 x - 1
9 + 1 = 2 x
log3 (2 x - 1) = 2
32 = 2 x - 1
9 = 2 x - 1
9 + 1 = 2 x
10 = 2 x
log3 (2 x - 1) = 2
32 = 2 x - 1
9 = 2 x - 1
9 + 1 = 2 x
10 = 2 x
= x10
2
log3 (2 x - 1) = 2
32 = 2 x - 1
9 = 2 x - 1
9 + 1 = 2 x
10 = 2 x
= x10
2
log3 (2 x - 1) = 2
Caso II
log (x + 6) = 1 + log (x – 3)
log (x + 6) = 1 + log (x – 3)
log (x + 6) - log (x – 3) = 1
log (x + 6) = 1 + log (x – 3)
log (x + 6) - log (x – 3) = 1
logx 6
x 3= 1
log (x + 6) = 1 + log (x – 3)
log (x + 6) - log (x – 3) = 1
logx 6
x 3= 1 101 =
x 6
x 3
log (x + 6) = 1 + log (x – 3)
log (x + 6) - log (x – 3) = 1
logx 6
x 3= 1 101 =
x 6
x 3
10 (x – 3) = x + 6
log (x + 6) = 1 + log (x – 3)
log (x + 6) - log (x – 3) = 1
logx 6
x 3= 1 101 =
x 6
x 3
10 (x – 3) = x + 6
10 x – 30 = x + 6
log (x + 6) = 1 + log (x – 3)
log (x + 6) - log (x – 3) = 1
logx 6
x 3= 1 101 =
x 6
x 3
10 (x – 3) = x + 6
10 x – 30 = x + 6
10 x – x = 30 + 6
log (x + 6) = 1 + log (x – 3)
log (x + 6) - log (x – 3) = 1
logx 6
x 3= 1 101 =
x 6
x 3
10 (x – 3) = x + 6
10 x – 30 = x + 6
10 x – x = 30 + 6
9 x = 36
log (x + 6) = 1 + log (x – 3)
log (x + 6) - log (x – 3) = 1
logx 6
x 3= 1 101 =
x 6
x 3
10 (x – 3) = x + 6
10 x – 30 = x + 6
10 x – x = 30 + 6
9 x = 36
36x9
log (x + 6) = 1 + log (x – 3)
log (x + 6) - log (x – 3) = 1
logx 6
x 3= 1 101 =
x 6
x 3
10 (x – 3) = x + 6
10 x – 30 = x + 6
10 x – x = 30 + 6
9 x = 36
36x9
Caso II (Continuación)
23 16lnx ln x lnx
3
23 16lnx ln x lnx
3
2
3
x 16ln lnx
3x
23 16lnx ln x lnx
3
2
3
x 16ln lnx
3x2
3
x.x 16ln
3x
23 16lnx ln x lnx
3
2
3
x 16ln lnx
3x2
3
x.x 16ln
3x3
1
3
x 16ln
3x
23 16lnx ln x lnx
3
2
3
x 16ln lnx
3x2
3
x.x 16ln
3x3
1
3
x 16ln
3x
8
316
ln x3
23 16lnx ln x lnx
3
2
3
x 16ln lnx
3x2
3
x.x 16ln
3x3
1
3
x 16ln
3x
8
316
ln x3
16 8
3 3e x
23 16lnx ln x lnx
3
2
3
x 16ln lnx
3x2
3
x.x 16ln
3x3
1
3
x 16ln
3x
8
316
ln x3
16 8
3 3e x
33 16 8e x
23 16lnx ln x lnx
3
2
3
x 16ln lnx
3x2
3
x.x 16ln
3x3
1
3
x 16ln
3x
8
316
ln x3
16 8
3 3e x
33 16 8e x
16 8e x
23 16lnx ln x lnx
3
2
3
x 16ln lnx
3x2
3
x.x 16ln
3x3
1
3
x 16ln
3x
8
316
ln x3
16 8
3 3e x
33 16 8e x
16 8e x
8 16e x2
23 16lnx ln x lnx
3
2
3
x 16ln lnx
3x2
3
x.x 16ln
3x3
1
3
x 16ln
3x
8
316
ln x3
16 8
3 3e x
33 16 8e x
16 8e x
8 16e x2
23 16lnx ln x lnx
3
23 16lnx ln x lnx
31
2316
lnx lnx lnx3
23 16lnx ln x lnx
31
2316
lnx lnx lnx3
1 16lnx .lnx 2.lnx
3 3
23 16lnx ln x lnx
31
2316
lnx lnx lnx3
1 16lnx .lnx 2.lnx
3 3
161lnx. 1 23 3
23 16lnx ln x lnx
31
2316
lnx lnx lnx3
1 16lnx .lnx 2.lnx
3 3
161lnx. 1 23 3
168lnx.3 3
23 16lnx ln x lnx
31
2316
lnx lnx lnx3
1 16lnx .lnx 2.lnx
3 3
161lnx. 1 23 3
168lnx.3 3
16 3lnx .
3 8
2
23 16lnx ln x lnx
31
2316
lnx lnx lnx3
1 16lnx .lnx 2.lnx
3 3
161lnx. 1 23 3
168lnx.3 3
16 3lnx .
3 8
ln x 2
2
23 16lnx ln x lnx
31
2316
lnx lnx lnx3
1 16lnx .lnx 2.lnx
3 3
161lnx. 1 23 3
168lnx.3 3
16 3lnx .
3 8
ln x 2
2
Caso III
(x + 1) + log3 (x + 1)2 = 8
2
3log
(x + 1) + log3 (x + 1)2 = 8
[log3 (x + 1)]2 + 2 log3 (x + 1) = 8
2
3log
(x + 1) + log3 (x + 1)2 = 8
| z = log3 (x + 1) |
[log3 (x + 1)]2 + 2 log3 (x + 1) = 8
2
3log
(x + 1) + log3 (x + 1)2 = 8
| z = log3 (x + 1) |
z2 + 2 z = 8
[log3 (x + 1)]2 + 2 log3 (x + 1) = 8
2
3log
(x + 1) + log3 (x + 1)2 = 8
| z = log3 (x + 1) |
z2 + 2 z = 8
z2 + 2 z – 8 = 0
[log3 (x + 1)]2 + 2 log3 (x + 1) = 8
2
3log
(x + 1) + log3 (x + 1)2 = 8
| z = log3 (x + 1) |
z2 + 2 z = 8
z2 + 2 z – 8 = 0
z1 = 2 z2 = – 4
[log3 (x + 1)]2 + 2 log3 (x + 1) = 8
2
3log
(x + 1) + log3 (x + 1)2 = 8
| z = log3 (x + 1) |
z2 + 2 z = 8
z2 + 2 z – 8 = 0
z1 = 2 z2 = – 4
[log3 (x + 1)]2 + 2 log3 (x + 1) = 8
z1 = log3 (x1 + 1) z2 = log3 (x2 + 1)
2
3log
(x + 1) + log3 (x + 1)2 = 8
| z = log3 (x + 1) |
z2 + 2 z = 8
z2 + 2 z – 8 = 0
z1 = 2 z2 = – 4
[log3 (x + 1)]2 + 2 log3 (x + 1) = 8
z1 = log3 (x1 + 1) z2 = log3 (x2 + 1)
2 = log3 (x1 + 1) – 4 = log3 (x2 + 1)
2
3log
(x + 1) + log3 (x + 1)2 = 8
| z = log3 (x + 1) |
z2 + 2 z = 8
z2 + 2 z – 8 = 0
z1 = 2 z2 = – 4
[log3 (x + 1)]2 + 2 log3 (x + 1) = 8
z1 = log3 (x1 + 1) z2 = log3 (x2 + 1)
2 = log3 (x1 + 1) – 4 = log3 (x2 + 1)
32 = x1 + 1
2
3log
(x + 1) + log3 (x + 1)2 = 8
| z = log3 (x + 1) |
z2 + 2 z = 8
z2 + 2 z – 8 = 0
z1 = 2 z2 = – 4
[log3 (x + 1)]2 + 2 log3 (x + 1) = 8
z1 = log3 (x1 + 1) z2 = log3 (x2 + 1)
2 = log3 (x1 + 1) – 4 = log3 (x2 + 1)
32 = x1 + 1
9 – 1 = x1
2
3log
(x + 1) + log3 (x + 1)2 = 8
| z = log3 (x + 1) |
z2 + 2 z = 8
z2 + 2 z – 8 = 0
z1 = 2 z2 = – 4
[log3 (x + 1)]2 + 2 log3 (x + 1) = 8
z1 = log3 (x1 + 1) z2 = log3 (x2 + 1)
2 = log3 (x1 + 1) – 4 = log3 (x2 + 1)
32 = x1 + 1
9 – 1 = x1
2
3log
(x + 1) + log3 (x + 1)2 = 8
| z = log3 (x + 1) |
z2 + 2 z = 8
z2 + 2 z – 8 = 0
z1 = 2 z2 = – 4
[log3 (x + 1)]2 + 2 log3 (x + 1) = 8
z1 = log3 (x1 + 1) z2 = log3 (x2 + 1)
2 = log3 (x1 + 1) – 4 = log3 (x2 + 1)
32 = x1 + 1 3 – 4 = x2 + 1
9 – 1 = x1
2
3log
(x + 1) + log3 (x + 1)2 = 8
| z = log3 (x + 1) |
z2 + 2 z = 8
z2 + 2 z – 8 = 0
z1 = 2 z2 = – 4
[log3 (x + 1)]2 + 2 log3 (x + 1) = 8
z1 = log3 (x1 + 1) z2 = log3 (x2 + 1)
2 = log3 (x1 + 1) – 4 = log3 (x2 + 1)
32 = x1 + 1 3 – 4 = x2 + 1
9 – 1 = x1– 1 = x2
181
2
3log
(x + 1) + log3 (x + 1)2 = 8
| z = log3 (x + 1) |
z2 + 2 z = 8
z2 + 2 z – 8 = 0
z1 = 2 z2 = – 4
[log3 (x + 1)]2 + 2 log3 (x + 1) = 8
z1 = log3 (x1 + 1) z2 = log3 (x2 + 1)
2 = log3 (x1 + 1) – 4 = log3 (x2 + 1)
32 = x1 + 1 3 – 4 = x2 + 1
9 – 1 = x1– 1 = x2
181
2
3log
Aquellas ecuaciones exponenciales que no se pueda expresar en términos de bases iguales, se utilizan los logaritmos y sus propiedades para hallar la solución.
