The document contains solutions to 25 logarithmic equations. It begins by stating the equations, then provides the step-by-step work and solution for each one. The solutions involve rewriting the logarithms in exponential form, isolating the variable, and evaluating.

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Ejercicios Resueltos
Ense˜nanza media
Logar´ıtmo
´Ultima edici´on: 10 de agosto de 2015
Determine el valor de x en las siguientes expresiones.
i) log2
1
8 = x
ii) log1
3
1
9 = x
iii) log2
5
8
125 = x
iv) log3
5
125
27 = x
v) log1
4
64 = x
vi) log1
2
16 = x
vii) log2(x − 3) = 1
viii) log5(2x + 3) = 3
ix) log3(x − 2) = −1
x) log2(3x − 2) = −2
xi) log2
1
x = 3
xii) log(x+1) 2 = 1
2
xiii) log(x−2) 3 = 1
3
xiv) log(x−1) 5 = −1
2
xv) log(x−2)
1
3 = −1
2
xvi) log3(x − 20) = −1
xvii) log2(3x − 1) = 5
3
xviii) log(x−5) 2 = 1
2
xix) log2(x − 6) = 1
xx) log√
x(2x − 5) = 2
xxi) log3(2 − x) = 1
2
xxii) log49
36
x = −1
2
xxiii) log4 x = 3
2
xxiv) logx 4 = −2
5
xxv) log16
1
2 = x

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Soluci´on:
i)
log2
1
8
= x,
2x
=
1
8
,
2x
=
1
2
3
,
2x
= 2−3
,
∴ x = −3.
ii)
log1
3
1
9
= x,
1
3
x
=
1
9
,
1
3
x
=
1
3
2
,
∴ x = 2.
iii)
log2
5
8
125
= x,
2
5
x
=
8
125
,
2
5
x
=
2
5
3
,
∴ x = 3.
iv)
log3
5
125
27
= x,
3
5
x
=
125
27
,
3
5
x
=
27
125
−1
,
3
5
x
=
3
5
−3
,
∴ x = −3.
2

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v)
log1
4
64 = x,
1
4
x
= 64,
1
4
x
=
1
64
−1
,
1
4
x
=
1
4
−3
,
∴ x = −3.
vi)
log1
2
16 = x,
1
2
x
= 16,
1
2
x
=
1
16
−1
,
1
2
x
=
1
2
−4
,
∴ x = −4.
vii)
log2(x − 3) = 1,
21
= x − 3,
2 + 3 = x,
∴ 5 = x.
viii)
log5(2x + 3) = 3,
53
= 2x + 3,
125 = 2x + 3,
125 − 3 = 2x,
122 = 2x,
122
2
= x,
∴ 61 = x.
3

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ix)
log3(x − 2) = −1,
3−1
= x − 2,
1
3
= x − 2,
1
3
+ 2 = x,
1 · 1 + 3 · 2
3
= x,
1 + 6
3
= x,
∴
7
3
= x.
x)
log2(3x − 2) = −2,
2−2
= 3x − 2,
1
2
2
= 3x − 2,
1
4
= 3x − 2,
1
4
+ 2 = 3x,
1 · 1 + 4 · 2
4
= 3x,
1 + 8
4
= 3x,
9
4
= 3x,
9
4 · 3
= x,
9
12
= x,
∴
3
4
= x.
xi)
log2
1
x
= 3,
23
=
1
x
,
8 =
1
x
,
8x = 1,
∴ x =
1
8
.
4

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xii)
log(x+1) 2 =
1
2
,
(x + 1)1/2
= 2,
√
x + 1 = 2, ( )2
√
x + 1
2
= 22
,
x + 1 = 4,
x = 4 − 1,
∴ x = 3.
xiii)
log(x−2) 3 =
1
3
,
(x − 2)1/3
= 3,
3√
x − 2 = 3, ( )3
3√
x − 2
3
= 33
,
x − 2 = 27,
x = 27 + 2,
∴ x = 29.
xiv)
log(x−1) 5 = −
1
2
,
(x − 1)−1/2
= 5,
1
√
x − 1
= 5, ( )2
1
√
x − 1
2
= 52
,
1
x − 1
= 25,
1 = 25 · (x − 1),
1
25
= x − 1,
1
25
+ 1 = x,
1 · 1 + 25 · 1
25
= x,
∴
26
25
= x.
5

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xv)
log(x−2)
1
3
= −
1
2
,
(x − 2)−1/2
=
1
3
,
1
√
x − 2
=
1
3
, ( )2
1
√
x − 2
2
=
1
3
2
,
1
x − 2
=
1
9
,
1 =
1
9
· (x − 2),
9 = x − 2,
9 + 2 = x,
∴ 11 = x.
xvi)
log3(x − 20) = −1,
3−1
= x − 20,
1
3
= x − 20,
1
3
+ 20 = x,
1 · 1 + 3 · 20
3
= x,
∴
61
3
= x.
xvii)
log2(3x − 1) =
5
3
,
25/3
= 3x − 1,
25/3
+ 1 = 3x,
25/3
+ 1
3
= x,
∴ 1.39 ≈ x.
6

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xviii)
log( x − 5)2 =
1
2
,
(x − 5)1/2
= 2,
(x − 5) = 2, ( )2
(x − 5)
2
= 22
,
x − 5 = 4,
x = 4 + 5,
∴ x = 9.
xix)
log2(x − 6) = 1,
21
= x − 6,
2 + 6 = x
∴ 8 = x.
xx)
log√
x(2x − 5) = 2,
√
x
2
= 2x − 5,
x = 2x − 5,
5 = 2x − x,
∴ 5 = x.
xxi)
log3(2 − x) =
1
2
,
31/2
= 2 − x,
x = 2 − 31/2
,
∴ x ≈ 0.28
xxii)
log49
36
x = −
1
2
,
49
36
−1/2
= x,
36
49
1/2
= x,
36
49
= x,
∴
6
7
= x.
7

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xxiii)
log4 x =
3
2
,
43/2
= x,
√
43 = x,
√
64 = x,
∴ 8 = x.
xxiv)
logx 4 = −
2
5
,
x−2/5
= 4,
5√
x−2 = 4, ( )5
5√
x−2
5
= 45
,
x−2
= 1024,
1
x2
= 1024,
1
1024
= x2
,
1
1024
= x,
∴
1
32
= x.
xxv)
log16
1
2
= x,
16x
=
1
2
,
24 x
= 2−1
,
24x
= 2−1
,
de aqu´ı deducimos que
4x = −1,
∴ x = −
1
4
.
8