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Fonction exponentielle de base $e$
Utiliser les propriétés algébriques de la fonction exponentielle - Exercice 4
1 min
0
Question 1
A
=
(
e
2
)
5
A=(e^{2})^5
A
=
(
e
2
)
5
Correction
(
e
a
)
b
=
e
a
×
b
\left(e^{a} \right)^{b} =e^{a\times b}
(
e
a
)
b
=
e
a
×
b
A
=
(
e
2
)
5
A=(e^{2})^5
A
=
(
e
2
)
5
équivaut successivement à :
A
=
e
2
×
5
A=e^{2\times5}
A
=
e
2
×
5
A
=
e
10
A=e^{10}
A
=
e
10
Question 2
B
=
(
e
6
)
4
B=(e^{6})^4
B
=
(
e
6
)
4
Correction
(
e
a
)
b
=
e
a
×
b
\left(e^{a} \right)^{b} =e^{a\times b}
(
e
a
)
b
=
e
a
×
b
B
=
(
e
6
)
4
B=(e^{6})^4
B
=
(
e
6
)
4
équivaut successivement à :
B
=
e
6
×
4
B=e^{6\times4}
B
=
e
6
×
4
B
=
e
24
B=e^{24}
B
=
e
24
Question 3
C
=
(
e
−
2
)
−
8
C=(e^{-2})^{-8 }
C
=
(
e
−
2
)
−
8
Correction
(
e
a
)
b
=
e
a
×
b
\left(e^{a} \right)^{b} =e^{a\times b}
(
e
a
)
b
=
e
a
×
b
C
=
(
e
−
2
)
−
8
C=(e^{-2})^{-8}
C
=
(
e
−
2
)
−
8
équivaut successivement à :
C
=
e
−
2
×
(
−
8
)
C=e^{-2\times(-8)}
C
=
e
−
2
×
(
−
8
)
C
=
e
16
C=e^{16}
C
=
e
16
Question 4
D
(
x
)
=
(
e
3
x
)
5
x
D(x)=(e^{3x})^{5x }
D
(
x
)
=
(
e
3
x
)
5
x
Correction
(
e
a
)
b
=
e
a
×
b
\left(e^{a} \right)^{b} =e^{a\times b}
(
e
a
)
b
=
e
a
×
b
D
(
x
)
=
(
e
3
x
)
5
x
D(x)=(e^{3x})^{5x}
D
(
x
)
=
(
e
3
x
)
5
x
équivaut successivement à :
D
(
x
)
=
e
3
x
×
(
5
x
)
D(x)=e^{3x\times(5x)}
D
(
x
)
=
e
3
x
×
(
5
x
)
D
(
x
)
=
e
15
x
2
D(x)=e^{15x^2}
D
(
x
)
=
e
15
x
2
Question 5
E
(
x
)
=
(
e
−
7
x
)
2
E(x)=(e^{-7x})^{2}
E
(
x
)
=
(
e
−
7
x
)
2
Correction
(
e
a
)
b
=
e
a
×
b
\left(e^{a} \right)^{b} =e^{a\times b}
(
e
a
)
b
=
e
a
×
b
E
(
x
)
=
(
e
−
7
x
)
2
E(x)=(e^{-7x})^{2}
E
(
x
)
=
(
e
−
7
x
)
2
équivaut successivement à :
E
(
x
)
=
e
−
7
x
×
2
E(x)=e^{-7x\times2}
E
(
x
)
=
e
−
7
x
×
2
E
(
x
)
=
e
−
14
x
E(x)=e^{-14x}
E
(
x
)
=
e
−
14
x
Question 6
F
(
x
)
=
(
e
2
)
x
+
1
F(x)=(e^{2})^{x+1}
F
(
x
)
=
(
e
2
)
x
+
1
Correction
(
e
a
)
b
=
e
a
×
b
\left(e^{a} \right)^{b} =e^{a\times b}
(
e
a
)
b
=
e
a
×
b
F
(
x
)
=
(
e
2
)
x
+
1
F(x)=(e^{2})^{x+1}
F
(
x
)
=
(
e
2
)
x
+
1
équivaut successivement à :
F
(
x
)
=
e
2
×
(
x
+
1
)
F(x)=e^{2\times(x+1)}
F
(
x
)
=
e
2
×
(
x
+
1
)
F
(
x
)
=
e
2
×
x
+
2
×
1
F(x)=e^{2\times{x}+2\times{1}}
F
(
x
)
=
e
2
×
x
+
2
×
1
F
(
x
)
=
e
2
x
+
2
F(x)=e^{2x+2}
F
(
x
)
=
e
2
x
+
2