Exponenciális egyenletek megoldása

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4^2x−1 ⋅ 2^x = 16^x

x = 2

4^2x = 1/4

x = −1/2

4^2x = −1/4

Egy pozizív szám (bal oldal) egyetlen hatványa sem lehet negatív. Tehát nincs megoldás!!!

3^3x = 0

Egy nem nulla szám egyetlen hatványa sem lehet 0. Tehát nincs megoldás!!!

0^5x = 0

A nulla akárhanyadik (kivéve 0) hatványa 0. Tehát az x bármi lehet a 0-t kivéve.

7^4x = 1

Bármely szám (kivéve a 0) nulladik hatványa 1. Tehát x = 0.

9^x−1 = 81 ⋅ √3

x = 13/4 = 3,25

0,5^4x+8 = 16

x = −3

(1/3)^2x+6 = 81

x = −5

10^x²−4x+3 = 1

x₁ = 1; x₂ = 3

2^3x+1 = 1/4

x = −1

10^x = 0,1 ⋅ √10

x = −1/2

0,5^3x−4 = 32

x = −1/3

6^2x² ⋅ 6^7x = 6¹⁵

x₁ = 3/2; x₂ = −5

27⋅2^x = 8⋅3^x

x = 3

9^5−2x = (1/3)^2x

x = 5

(23/42)^2−x² − 1 = 0

x₁ = √2; x₂ = −√2

125⋅3^x−1 = 3⋅5^x+1

x = 2

(2/7)^3x−7 = (7/2)^7x−3

x = 1

2 ⋅ 3^x+2 − 42 ⋅ 3^x−1 = 12

x = 1

2^x−1 + 2^x+1 = 20

x = 3

3 ⋅ 3^x+3 − 3^x+2 + 3 ⋅ 3^x = 25

x = −1

2^3x²−1 = 4^x

x₁ = 1; x₂ = −1/3

3^x + 3^x+2 + 3^x−1 = 31/3

x = 0

2^x+3 − 2^x−2 + 2^x+1 = 39

x = 2

25^x + 5 = 6 ⋅ 5^x

x₁ = 0; x₂ = 1

(2/3)^x ⋅ (9/8)^x = 27/64

x = 3

9^x + 6⋅3^x − 27 = 0

x = 1

5^5+x + 5^x+6 − 8 = 142

x = −3

2^x+2 + 4^x−1 = 128

x = 4

3^2x−5 + 3^x−2 − 3 = 33

x = 4

5^x−1 ⋅ 125^x+2 = 25^x+1

x = −3/2

3^x+2 ⋅ 9^x = 27^x−2

x = 4/3

10^x+3 ⋅ 100^5+2x = 0,1^x

x = −13/4

10^2−x ⋅ 100^5+2x = 0,1^x

x = −3

2^1+x + 2^x+2 = 2^x+3 − 40 + 2^x−1

x = 4