EQUACOES EXPONENCIAIS Questoes Resolvidas

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Exercícios Resolvidos 
EQUAÇÕES EXPONENCIAIS 
𝑎) 2𝑥+1 = 23 ⇒ 𝑥 + 1 = 3 ∴ 𝑥 = 2 
𝑏) 5𝑥
2+2𝑥 = 53 ⇒ 𝑥2 + 2𝑥 = 3 
𝑥2 + 2𝑥 − 3 = 0 ∴ 𝑥 = −3 𝑜𝑢 𝑥 = 1 
𝑐) 34−2𝑥 = 32 2𝑥 
34−2𝑥 = 34𝑥 ⇒ 4 − 2𝑥 = 4𝑥 
4 = 6𝑥 ∴ 𝑥 =
 4 
6
=
 2 
3
 
𝑑) 22 𝑥 = 23 
22𝑥 = 23 ⇒ 2𝑥 = 3 ∴ 𝑥 =
 3 
2
 
𝑒) 23 𝑥+1 = 25 
23𝑥+3 = 25 ⇒ 3𝑥 + 3 = 5 
3𝑥 = 2 ∴ 𝑥 =
 2 
3
 
𝑓) 3𝑥
2+𝑥 = 32 ⇒ 𝑥2 + 𝑥 = 2 
𝑥2 + 𝑥 − 2 = 0 ∴ 𝑥 = −2 𝑜𝑢 𝑥 = 1 
𝑔) 3𝑥 = 34 ⇒ 𝑥 = 4 
𝑕) 2𝑥−3 = 24 ⇒ 𝑥 − 3 = 4 ∴ 𝑥 = 7 
𝑖) 4𝑥 = 43 ⇒ 𝑥 = 3 
𝑗) 2𝑥+1 = 20 𝑗) 2𝑥+1 = 20 ⇒ 𝑥 + 1 = 0 ∴ 𝑥 = −1 
𝑘) 23 𝑥+1 = 27 𝑥−3 
23𝑥+3 = 27𝑥−21 ⇒ 3𝑥 + 3 = 7𝑥 − 21 
21 + 3 = 7𝑥 − 3𝑥 
4𝑥 = 24 ∴ 𝑥 = 6 
𝑙) 25𝑥+2 = 250 𝑙) 25𝑥+2 = 250 ⇒ 𝑥 + 2 = 0 ∴ 𝑥 = −2 
𝑎) 4−1 𝑥−2 = 42 𝑥 
4−𝑥+2 = 42𝑥 ⇒ −𝑥 + 2 = 2𝑥 
2 = 3𝑥 ∴ 𝑥 =
 2 
3
 
𝑏) 34 𝑥
2−𝑥 = 3−3 −8 
34𝑥
2−4𝑥 = 324 ⇒ 4𝑥2 − 4𝑥 = 24 
4𝑥2 − 4𝑥 − 24 = 0 
𝑥2 − 𝑥 − 6 = 0 ∴ 𝑥 = −2 𝑜𝑢 𝑥 = 3 
𝑐) 10𝑥
2
= 104 ⇒ 𝑥2 = 4 ∴ 𝑥 = ± 4 = ±2 
𝑑) 2𝑥 = 2−5 ⇒ 𝑥 = −5 
𝑒) 32 𝑥 = 3−1 
32𝑥 = 3−1 ⇒ 2𝑥 = −1 ∴ 𝑥 = −
 1 
2
 
𝑓) 5𝑥 = 5−4 ⇒ 𝑥 = −4 
𝑎) 51
2 𝑥
= 53 
5
1
2 
𝑥
= 53 
5
𝑥
2 = 53 ⇒ 
 𝑥 
2
= 3 ∴ 𝑥 = 6 
𝑏) 6𝑥
2
= 62 3 
6
𝑥
2 = 66 ⇒ 
 𝑥 
2
= 6 ∴ 𝑥 = 12 
𝑐) 5𝑥 = 51
2
 
5𝑥 = 5
1
2 ⇒ 𝑥 =
 1 
2
 
𝑑) 72 𝑥 = 71
2
 
72𝑥 = 7
1
2 ⇒ 2𝑥 =
 1 
2
 ∴ 𝑥 =
 1 
4
 
𝑒) 25 𝑥+3 = 21
3
 
25𝑥+15 = 2
1
3 
5𝑥 + 15 =
 1 
3
 
15𝑥 + 45 = 1 
15𝑥 = −44 ∴ 𝑥 = −
44
15
 
𝑎) 4𝑥 . 41 −
 4𝑥
41
= 60 
4𝑥 . 4 −
 4𝑥
4
= 60 
4𝑥 = 𝑦 
4𝑦 −
𝑦
4
= 60 
16𝑦 − 𝑦 = 60 × 4 
15𝑦 = 60 × 4 
𝑦 =
60 × 4
15
 
1 
4 
𝑦 = 16 𝑦 = 16 
4𝑥 = 𝑦 
𝑦 = 16 
⇓ 
4𝑥 = 16 
4𝑥 = 42 
∴ 𝑥 = 2 
𝑏) 
5𝑥
51
+
5𝑥
52
= 6 
5𝑥
5
+
5𝑥
25
= 6 
5𝑥 = 𝑦 
 𝑦 
5
+
𝑦
25
= 6 
5𝑦 + 𝑦 = 6 × 25 
6𝑦 = 6 × 25 
1 1 
𝑦 = 25 
5𝑥 = 𝑦 
𝑦 = 25 
⇓ 
5𝑥 = 25 
5𝑥 = 52 
∴ 𝑥 = 2 
𝑐) 3𝑥. 31 + 3𝑥 = 36 
3𝑥. 3 + 3𝑥 = 36 
3𝑥 = 𝑦 
3𝑦 + 𝑦 = 36 
4𝑦 = 36 
𝑦 = 9 
⇓ 
3𝑥 = 9 
3𝑥 = 32 
∴ 𝑥 = 2 
 𝑑) 7𝑥 . 71 − 49 .
7𝑥
72
= 42 
 7𝑥 . 7 − 49 .
7𝑥
49
= 42 
7𝑥 = 𝑦 
7𝑦 − 𝑦 = 42 
6𝑦 = 42 
𝑦 = 7 
⇓ 
7𝑥 = 7 
7𝑥 = 71 
∴ 𝑥 = 1 
𝑒) 3𝑥 . 31 + 3𝑥. 32 = 12 
3𝑥 = 𝑦 
3𝑦 + 9𝑦 = 12 
12𝑦 = 12 
𝑦 = 1 
⇓ 
3𝑥 = 1 
3𝑥 = 30 
∴ 𝑥 = 0 
𝑎) 22 𝑥 − 6. 2𝑥 + 8 = 0 
2𝑥 2 − 6. 2𝑥 + 8 = 0 
2𝑥 = 𝑦 
𝑦2 − 6𝑦 + 8 = 0 
𝑦 = 2 𝑜𝑢 𝑦 = 4 
𝑆𝑒 𝑦 = 2 ⇒ 2𝑥 = 2 ∴ 𝑥 = 1 
𝑆𝑒 𝑦 = 4 ⇒ 2𝑥 = 4 ∴ 𝑥 = 2 
𝑅𝑒𝑠𝑝𝑜𝑠𝑡𝑎: 𝑥 = 1 𝑜𝑢 𝑥 = 2 
𝑏) 52 𝑥 − 2. 5𝑥 + 1 = 0 
5𝑥 2 − 2. 5𝑥 + 1 = 0 
5𝑥 = 𝑦 
𝑦2 − 2𝑦 + 1 = 0 
𝑦 = 1 
 𝑦 = 1 ⇒ 5𝑥 = 1 ∴ 𝑥 = 0 
𝑅𝑒𝑠𝑝𝑜𝑠𝑡𝑎: 𝑥 = 0 
𝑐) 32 𝑥 + 3 .
3𝑥
31
= 12 
3𝑥 2 + 3 .
3𝑥
3
− 12 = 0 
3𝑥 = 𝑦 
𝑦2 + 𝑦 − 12 = 0 
𝑦 = −4 𝑜𝑢 𝑦 = 3 
𝑆𝑒 𝑦 = −4 ⇒ 3𝑥 = −4 ∴ ∄ 𝑥 ∈ ℝ 
𝑆𝑒 𝑦 = 3 ⇒ 3𝑥 = 3 ∴ 𝑥 = 1 
𝑅𝑒𝑠𝑝𝑜𝑠𝑡𝑎: 𝑥 = 1 
𝑑) 22 𝑥 − 3. 2𝑥 + 2 = 0 
2𝑥 2 − 3. 2𝑥 + 2 = 0 
2𝑥 = 𝑦 
𝑦2 − 3𝑦 + 2 = 0 
𝑦 = 1 𝑜𝑢 𝑦 = 2 
𝑆𝑒 𝑦 = 1 ⇒ 2𝑥 = 1 ∴ 𝑥 = 0 
𝑆𝑒 𝑦 = 2 ⇒ 2𝑥 = 2 ∴ 𝑥 = 1 
𝑅𝑒𝑠𝑝𝑜𝑠𝑡𝑎: 𝑥 = 0 𝑜𝑢 𝑥 = 1 
3𝑥
2−3𝑥 = 3−2 ⇒ 𝑥2 − 3𝑥 = −2 
𝑥2 − 3𝑥 + 2 = 0 ∴ 𝑥 = 1 𝑜𝑢 𝑥 = 2 
2𝑥 − 2𝑥 . 21 + 8 = 0 
2𝑥 = 𝑦 
𝑦 − 2𝑦 = −8 
−𝑦 = −8 ∴ 𝑦 = 8 
2𝑥 = 8 ⇒ 2𝑥 = 23 ∴ 𝑥 = 3 
2𝑥+1
2
=
25
100
 
2
𝑥+1
2 =
 1 
4
 =
1
22
 
2
𝑥+1
2 = 2−2 ⇒ 
𝑥 + 1
2
= −2 
𝑥 + 1 = −4 
𝑥 = −5 
1 
4 
= 2−2 
2𝑥 . 22 𝑥+1 . 23 𝑥+2 = 24 𝑥+3 
2𝑥 . 22𝑥+2 . 23𝑥+6 = 24𝑥+12 
2𝑥+2𝑥+2+3𝑥+6 = 24𝑥+12 
26𝑥+8 = 24𝑥+12 ⇒ 6𝑥 + 8 = 4𝑥 + 12 
2𝑥 = 4 ∴ 𝑥 = 2 
2𝑥+3 . 16 = 641
2
 
2𝑥+3 . 16 = 8 
2𝑥+3 =
8
16
 
1 
2 
=
 1 
2
 = 2−1 
2𝑥+3 = 2−1 ⇒ 𝑥 + 3 = −1 
∴ 𝑥 = −4 
22 𝑥 = 25 
22𝑥 = 25 ⇒ 2𝑥 = 5 ∴ 𝑥 = 2,5 
2
45
𝑥 = 25 ⇒
45
𝑥
= 5 ∴ 𝑥 = 9 
72 2−𝑥 = 7−1 
74−2𝑥 = 7−1 ⇒ 4 − 2𝑥 = −1 
4 + 1 = 2𝑥 
2𝑥 = 5 ∴ 𝑥 = 2,5 
ISERJ – 2012 
Professora Telma Castro Silva 
Fonte: http://blog.educacional.com.br/mjustina/