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La fonction exponentielle
Propriétés algébriques de la fonction exponentielle - Exercice 1
25 min
40
Simplifier les expressions suivantes :
Question 1
a
(
x
)
=
e
3
e
4
a\left(x\right)=e^{3} e^{4}
a
(
x
)
=
e
3
e
4
Correction
e
a
e
b
=
e
a
+
b
e^{a} e^{b} =e^{a+b}
e
a
e
b
=
e
a
+
b
e
a
e
b
=
e
a
−
b
\frac{e^{a} }{e^{b} } =e^{a-b}
e
b
e
a
=
e
a
−
b
(
e
a
)
b
=
e
a
×
b
\left(e^{a} \right)^{b} =e^{a\times b}
(
e
a
)
b
=
e
a
×
b
e
−
a
=
1
e
a
e^{-a} =\frac{1}{e^{a} }
e
−
a
=
e
a
1
a
(
x
)
=
e
3
e
4
a\left(x\right)=e^{3} e^{4}
a
(
x
)
=
e
3
e
4
équivaut successivement à
a
(
x
)
=
e
3
+
4
a\left(x\right)=e^{3+4}
a
(
x
)
=
e
3
+
4
a
(
x
)
=
e
7
a\left(x\right)=e^{7}
a
(
x
)
=
e
7
Question 2
b
(
x
)
=
e
−
5
e
2
b\left(x\right)=\frac{e^{-5} }{e^{2} }
b
(
x
)
=
e
2
e
−
5
Correction
e
a
e
b
=
e
a
+
b
e^{a} e^{b} =e^{a+b}
e
a
e
b
=
e
a
+
b
e
a
e
b
=
e
a
−
b
\frac{e^{a} }{e^{b} } =e^{a-b}
e
b
e
a
=
e
a
−
b
(
e
a
)
b
=
e
a
×
b
\left(e^{a} \right)^{b} =e^{a\times b}
(
e
a
)
b
=
e
a
×
b
e
−
a
=
1
e
a
e^{-a} =\frac{1}{e^{a} }
e
−
a
=
e
a
1
b
(
x
)
=
e
−
5
e
2
b\left(x\right)=\frac{e^{-5} }{e^{2} }
b
(
x
)
=
e
2
e
−
5
équivaut successivement à
b
(
x
)
=
e
−
5
−
2
b\left(x\right)=e^{-5-2}
b
(
x
)
=
e
−
5
−
2
b
(
x
)
=
e
−
7
b\left(x\right)=e^{-7}
b
(
x
)
=
e
−
7
Question 3
c
(
x
)
=
(
e
−
5
)
2
e
2
e
−
6
c\left(x\right)=\frac{\left(e^{-5} \right)^{2} }{e^{2} e^{-6} }
c
(
x
)
=
e
2
e
−
6
(
e
−
5
)
2
Correction
e
a
e
b
=
e
a
+
b
e^{a} e^{b} =e^{a+b}
e
a
e
b
=
e
a
+
b
e
a
e
b
=
e
a
−
b
\frac{e^{a} }{e^{b} } =e^{a-b}
e
b
e
a
=
e
a
−
b
(
e
a
)
b
=
e
a
×
b
\left(e^{a} \right)^{b} =e^{a\times b}
(
e
a
)
b
=
e
a
×
b
e
−
a
=
1
e
a
e^{-a} =\frac{1}{e^{a} }
e
−
a
=
e
a
1
c
(
x
)
=
(
e
−
5
)
2
e
2
e
−
6
c\left(x\right)=\frac{\left(e^{-5} \right)^{2} }{e^{2} e^{-6} }
c
(
x
)
=
e
2
e
−
6
(
e
−
5
)
2
équivaut successivement à
c
(
x
)
=
e
−
5
×
2
e
2
+
(
−
6
)
c\left(x\right)=\frac{e^{-5\times 2} }{e^{2+\left(-6\right)} }
c
(
x
)
=
e
2
+
(
−
6
)
e
−
5
×
2
c
(
x
)
=
e
−
10
e
−
4
c\left(x\right)=\frac{e^{-10} }{e^{-4} }
c
(
x
)
=
e
−
4
e
−
10
c
(
x
)
=
e
−
10
−
(
−
4
)
c\left(x\right)=e^{-10-\left(-4\right)}
c
(
x
)
=
e
−
10
−
(
−
4
)
c
(
x
)
=
e
−
6
c\left(x\right)=e^{-6}
c
(
x
)
=
e
−
6
Question 4
d
(
x
)
=
e
−
4
(
e
−
5
)
2
(
e
2
)
5
e
−
6
d\left(x\right)=\frac{e^{-4} \left(e^{-5} \right)^{2} }{\left(e^{2} \right)^{5} e^{-6} }
d
(
x
)
=
(
e
2
)
5
e
−
6
e
−
4
(
e
−
5
)
2
Correction
e
a
e
b
=
e
a
+
b
e^{a} e^{b} =e^{a+b}
e
a
e
b
=
e
a
+
b
e
a
e
b
=
e
a
−
b
\frac{e^{a} }{e^{b} } =e^{a-b}
e
b
e
a
=
e
a
−
b
(
e
a
)
b
=
e
a
×
b
\left(e^{a} \right)^{b} =e^{a\times b}
(
e
a
)
b
=
e
a
×
b
e
−
a
=
1
e
a
e^{-a} =\frac{1}{e^{a} }
e
−
a
=
e
a
1
d
(
x
)
=
e
−
4
(
e
−
5
)
2
(
e
2
)
5
e
−
6
d\left(x\right)=\frac{e^{-4} \left(e^{-5} \right)^{2} }{\left(e^{2} \right)^{5} e^{-6} }
d
(
x
)
=
(
e
2
)
5
e
−
6
e
−
4
(
e
−
5
)
2
équivaut successivement à
d
(
x
)
=
e
−
4
e
−
5
×
2
e
2
×
5
e
−
6
d\left(x\right)=\frac{e^{-4} e^{-5\times 2} }{e^{2\times 5} e^{-6} }
d
(
x
)
=
e
2
×
5
e
−
6
e
−
4
e
−
5
×
2
d
(
x
)
=
e
−
4
e
−
10
e
10
e
−
6
d\left(x\right)=\frac{e^{-4} e^{-10} }{e^{10} e^{-6} }
d
(
x
)
=
e
10
e
−
6
e
−
4
e
−
10
d
(
x
)
=
e
−
4
+
(
−
10
)
e
10
+
(
−
6
)
d\left(x\right)=\frac{e^{-4+\left(-10\right)} }{e^{10+\left(-6\right)} }
d
(
x
)
=
e
10
+
(
−
6
)
e
−
4
+
(
−
10
)
d
(
x
)
=
e
−
14
e
4
d\left(x\right)=\frac{e^{-14} }{e^{4} }
d
(
x
)
=
e
4
e
−
14
d
(
x
)
=
e
−
14
−
4
d\left(x\right)=e^{-14-4}
d
(
x
)
=
e
−
14
−
4
d
(
x
)
=
e
−
18
d\left(x\right)=e^{-18}
d
(
x
)
=
e
−
18
Question 5
f
(
x
)
=
e
2
x
+
1
e
−
3
x
+
5
f\left(x\right)=e^{2x+1} e^{-3x+5}
f
(
x
)
=
e
2
x
+
1
e
−
3
x
+
5
Correction
e
a
e
b
=
e
a
+
b
e^{a} e^{b} =e^{a+b}
e
a
e
b
=
e
a
+
b
e
a
e
b
=
e
a
−
b
\frac{e^{a} }{e^{b} } =e^{a-b}
e
b
e
a
=
e
a
−
b
(
e
a
)
b
=
e
a
×
b
\left(e^{a} \right)^{b} =e^{a\times b}
(
e
a
)
b
=
e
a
×
b
e
−
a
=
1
e
a
e^{-a} =\frac{1}{e^{a} }
e
−
a
=
e
a
1
f
(
x
)
=
e
2
x
+
1
e
−
3
x
+
5
f\left(x\right)=e^{2x+1} e^{-3x+5}
f
(
x
)
=
e
2
x
+
1
e
−
3
x
+
5
équivaut successivement à
f
(
x
)
=
e
2
x
+
1
+
(
−
3
x
+
5
)
f\left(x\right)=e^{2x+1+\left(-3x+5\right)}
f
(
x
)
=
e
2
x
+
1
+
(
−
3
x
+
5
)
f
(
x
)
=
e
−
x
+
6
f\left(x\right)=e^{-x+6}
f
(
x
)
=
e
−
x
+
6
Question 6
g
(
x
)
=
e
−
x
+
1
e
3
x
−
4
g\left(x\right)=\frac{e^{-x+1} }{e^{3x-4} }
g
(
x
)
=
e
3
x
−
4
e
−
x
+
1
Correction
e
a
e
b
=
e
a
+
b
e^{a} e^{b} =e^{a+b}
e
a
e
b
=
e
a
+
b
e
a
e
b
=
e
a
−
b
\frac{e^{a} }{e^{b} } =e^{a-b}
e
b
e
a
=
e
a
−
b
(
e
a
)
b
=
e
a
×
b
\left(e^{a} \right)^{b} =e^{a\times b}
(
e
a
)
b
=
e
a
×
b
e
−
a
=
1
e
a
e^{-a} =\frac{1}{e^{a} }
e
−
a
=
e
a
1
g
(
x
)
=
e
−
x
+
1
e
3
x
−
4
g\left(x\right)=\frac{e^{-x+1} }{e^{3x-4} }
g
(
x
)
=
e
3
x
−
4
e
−
x
+
1
équivaut successivement à
g
(
x
)
=
e
−
x
+
1
−
(
3
x
−
4
)
g\left(x\right)=e^{-x+1-\left(3x-4\right)}
g
(
x
)
=
e
−
x
+
1
−
(
3
x
−
4
)
g
(
x
)
=
e
−
x
+
1
−
3
x
+
4
g\left(x\right)=e^{-x+1-3x+4}
g
(
x
)
=
e
−
x
+
1
−
3
x
+
4
g
(
x
)
=
e
−
4
x
+
5
g\left(x\right)=e^{-4x+5}
g
(
x
)
=
e
−
4
x
+
5
Question 7
h
(
x
)
=
(
e
3
x
+
2
)
2
h\left(x\right)=\left(e^{3x+2} \right)^{2}
h
(
x
)
=
(
e
3
x
+
2
)
2
Correction
e
a
e
b
=
e
a
+
b
e^{a} e^{b} =e^{a+b}
e
a
e
b
=
e
a
+
b
e
a
e
b
=
e
a
−
b
\frac{e^{a} }{e^{b} } =e^{a-b}
e
b
e
a
=
e
a
−
b
(
e
a
)
b
=
e
a
×
b
\left(e^{a} \right)^{b} =e^{a\times b}
(
e
a
)
b
=
e
a
×
b
e
−
a
=
1
e
a
e^{-a} =\frac{1}{e^{a} }
e
−
a
=
e
a
1
h
(
x
)
=
(
e
3
x
+
2
)
2
h\left(x\right)=\left(e^{3x+2} \right)^{2}
h
(
x
)
=
(
e
3
x
+
2
)
2
équivaut successivement à
h
(
x
)
=
e
(
3
x
+
2
)
×
2
h\left(x\right)=e^{\left(3x+2\right)\times2}
h
(
x
)
=
e
(
3
x
+
2
)
×
2
h
(
x
)
=
e
3
x
×
2
+
2
×
2
h\left(x\right)=e^{3x\times 2+2\times 2}
h
(
x
)
=
e
3
x
×
2
+
2
×
2
h
(
x
)
=
e
6
x
+
4
h\left(x\right)=e^{6x+4}
h
(
x
)
=
e
6
x
+
4
Question 8
i
(
x
)
=
e
5
x
+
7
e
−
x
−
3
e
2
x
+
3
i\left(x\right)=\frac{e^{5x+7} e^{-x-3} }{e^{2x+3} }
i
(
x
)
=
e
2
x
+
3
e
5
x
+
7
e
−
x
−
3
Correction
e
a
e
b
=
e
a
+
b
e^{a} e^{b} =e^{a+b}
e
a
e
b
=
e
a
+
b
e
a
e
b
=
e
a
−
b
\frac{e^{a} }{e^{b} } =e^{a-b}
e
b
e
a
=
e
a
−
b
(
e
a
)
b
=
e
a
×
b
\left(e^{a} \right)^{b} =e^{a\times b}
(
e
a
)
b
=
e
a
×
b
e
−
a
=
1
e
a
e^{-a} =\frac{1}{e^{a} }
e
−
a
=
e
a
1
i
(
x
)
=
e
5
x
+
7
e
−
x
−
3
e
2
x
+
3
i\left(x\right)=\frac{e^{5x+7} e^{-x-3} }{e^{2x+3} }
i
(
x
)
=
e
2
x
+
3
e
5
x
+
7
e
−
x
−
3
équivaut successivement à
i
(
x
)
=
e
5
x
+
7
+
(
−
x
−
3
)
e
2
x
+
3
i\left(x\right)=\frac{e^{5x+7+\left(-x-3\right)} }{e^{2x+3} }
i
(
x
)
=
e
2
x
+
3
e
5
x
+
7
+
(
−
x
−
3
)
i
(
x
)
=
e
4
x
+
4
e
2
x
+
3
i\left(x\right)=\frac{e^{4x+4} }{e^{2x+3} }
i
(
x
)
=
e
2
x
+
3
e
4
x
+
4
i
(
x
)
=
e
4
x
+
4
−
(
2
x
+
3
)
i\left(x\right)=e^{4x+4-\left(2x+3\right)}
i
(
x
)
=
e
4
x
+
4
−
(
2
x
+
3
)
i
(
x
)
=
e
4
x
+
4
−
2
x
−
3
i\left(x\right)=e^{4x+4-2x-3}
i
(
x
)
=
e
4
x
+
4
−
2
x
−
3
i
(
x
)
=
e
2
x
+
1
i\left(x\right)=e^{2x+1}
i
(
x
)
=
e
2
x
+
1
Question 9
j
(
x
)
=
e
−
2
x
+
6
(
e
4
x
+
1
)
3
e
−
x
+
4
j\left(x\right)=\frac{e^{-2x+6}\left(e^{4x+1} \right)^{3} }{e^{-x+4} }
j
(
x
)
=
e
−
x
+
4
e
−
2
x
+
6
(
e
4
x
+
1
)
3
Correction
e
a
e
b
=
e
a
+
b
e^{a} e^{b} =e^{a+b}
e
a
e
b
=
e
a
+
b
e
a
e
b
=
e
a
−
b
\frac{e^{a} }{e^{b} } =e^{a-b}
e
b
e
a
=
e
a
−
b
(
e
a
)
b
=
e
a
×
b
\left(e^{a} \right)^{b} =e^{a\times b}
(
e
a
)
b
=
e
a
×
b
e
−
a
=
1
e
a
e^{-a} =\frac{1}{e^{a} }
e
−
a
=
e
a
1
j
(
x
)
=
e
−
2
x
+
6
(
e
4
x
+
1
)
3
e
−
x
+
4
j\left(x\right)=\frac{e^{-2x+6}\left(e^{4x+1} \right)^{3} }{e^{-x+4} }
j
(
x
)
=
e
−
x
+
4
e
−
2
x
+
6
(
e
4
x
+
1
)
3
équivaut successivement à
j
(
x
)
=
e
−
2
x
+
6
×
e
(
4
x
+
1
)
×
3
e
−
x
+
4
j\left(x\right)=\frac{e^{-2x+6} \times e^{\left(4x+1\right)\times 3} }{e^{-x+4} }
j
(
x
)
=
e
−
x
+
4
e
−
2
x
+
6
×
e
(
4
x
+
1
)
×
3
j
(
x
)
=
e
−
2
x
+
6
×
e
4
x
×
3
+
1
×
3
e
−
x
+
4
j\left(x\right)=\frac{e^{-2x+6} \times e^{4x\times 3+1\times 3} }{e^{-x+4} }
j
(
x
)
=
e
−
x
+
4
e
−
2
x
+
6
×
e
4
x
×
3
+
1
×
3
j
(
x
)
=
e
−
2
x
+
6
e
12
x
+
3
e
−
x
+
4
j\left(x\right)=\frac{e^{-2x+6} e^{12x+3} }{e^{-x+4} }
j
(
x
)
=
e
−
x
+
4
e
−
2
x
+
6
e
12
x
+
3
j
(
x
)
=
e
−
2
x
+
6
+
12
x
+
3
e
−
x
+
4
j\left(x\right)=\frac{e^{-2x+6+12x+3} }{e^{-x+4} }
j
(
x
)
=
e
−
x
+
4
e
−
2
x
+
6
+
12
x
+
3
j
(
x
)
=
e
10
x
+
9
e
−
x
+
4
j\left(x\right)=\frac{e^{10x+9} }{e^{-x+4} }
j
(
x
)
=
e
−
x
+
4
e
10
x
+
9
j
(
x
)
=
e
10
x
+
9
−
(
−
x
+
4
)
j\left(x\right)=e^{10x+9-\left(-x+4\right)}
j
(
x
)
=
e
10
x
+
9
−
(
−
x
+
4
)
j
(
x
)
=
e
10
x
+
9
+
x
−
4
j\left(x\right)=e^{10x+9+x-4}
j
(
x
)
=
e
10
x
+
9
+
x
−
4
j
(
x
)
=
e
11
x
+
5
j\left(x\right)=e^{11x+5}
j
(
x
)
=
e
11
x
+
5
Question 10
k
(
x
)
=
(
e
2
x
2
−
x
e
−
x
+
4
)
2
k\left(x\right)=\left(\frac{e^{2x^{2} -x} }{e^{-x+4} } \right)^{2}
k
(
x
)
=
(
e
−
x
+
4
e
2
x
2
−
x
)
2
Correction
e
a
e
b
=
e
a
+
b
e^{a} e^{b} =e^{a+b}
e
a
e
b
=
e
a
+
b
e
a
e
b
=
e
a
−
b
\frac{e^{a} }{e^{b} } =e^{a-b}
e
b
e
a
=
e
a
−
b
(
e
a
)
b
=
e
a
×
b
\left(e^{a} \right)^{b} =e^{a\times b}
(
e
a
)
b
=
e
a
×
b
e
−
a
=
1
e
a
e^{-a} =\frac{1}{e^{a} }
e
−
a
=
e
a
1
k
(
x
)
=
(
e
2
x
2
−
x
e
−
x
+
4
)
2
k\left(x\right)=\left(\frac{e^{2x^{2} -x} }{e^{-x+4} } \right)^{2}
k
(
x
)
=
(
e
−
x
+
4
e
2
x
2
−
x
)
2
équivaut successivement à :
k
(
x
)
=
(
e
2
x
2
−
x
)
2
(
e
−
x
+
4
)
2
k\left(x\right)=\frac{\left(e^{2x^{2} -x} \right)^{2} }{\left(e^{-x+4} \right)^{2} }
k
(
x
)
=
(
e
−
x
+
4
)
2
(
e
2
x
2
−
x
)
2
k
(
x
)
=
e
(
2
x
2
−
x
)
×
2
e
(
−
x
+
4
)
×
2
k\left(x\right)=\frac{e^{\left(2x^{2} -x\right)\times2} }{e^{\left(-x+4\right)\times2} }
k
(
x
)
=
e
(
−
x
+
4
)
×
2
e
(
2
x
2
−
x
)
×
2
k
(
x
)
=
e
4
x
2
−
2
x
e
−
2
x
+
8
k\left(x\right)=\frac{e^{4x^{2} -2x} }{e^{-2x+8} }
k
(
x
)
=
e
−
2
x
+
8
e
4
x
2
−
2
x
k
(
x
)
=
e
4
x
2
−
2
x
−
(
−
2
x
+
8
)
k\left(x\right)=e^{4x^{2} -2x-\left(-2x+8\right)}
k
(
x
)
=
e
4
x
2
−
2
x
−
(
−
2
x
+
8
)
k
(
x
)
=
e
4
x
2
−
2
x
+
2
x
−
8
k\left(x\right)=e^{4x^{2} -2x+2x-8}
k
(
x
)
=
e
4
x
2
−
2
x
+
2
x
−
8
k
(
x
)
=
e
4
x
2
−
8
k\left(x\right)=e^{4x^{2} -8}
k
(
x
)
=
e
4
x
2
−
8
Question 11
l
(
x
)
=
e
x
−
7
e
2
x
×
e
3
x
+
5
e
−
2
x
+
1
l\left(x\right)=\frac{e^{x-7} }{e^{2x} } \times \frac{e^{3x+5} }{e^{-2x+1} }
l
(
x
)
=
e
2
x
e
x
−
7
×
e
−
2
x
+
1
e
3
x
+
5
Correction
e
a
e
b
=
e
a
+
b
e^{a} e^{b} =e^{a+b}
e
a
e
b
=
e
a
+
b
e
a
e
b
=
e
a
−
b
\frac{e^{a} }{e^{b} } =e^{a-b}
e
b
e
a
=
e
a
−
b
(
e
a
)
b
=
e
a
×
b
\left(e^{a} \right)^{b} =e^{a\times b}
(
e
a
)
b
=
e
a
×
b
e
−
a
=
1
e
a
e^{-a} =\frac{1}{e^{a} }
e
−
a
=
e
a
1
l
(
x
)
=
e
x
−
7
e
2
x
×
e
3
x
+
5
e
−
2
x
+
1
l\left(x\right)=\frac{e^{x-7} }{e^{2x} } \times \frac{e^{3x+5} }{e^{-2x+1} }
l
(
x
)
=
e
2
x
e
x
−
7
×
e
−
2
x
+
1
e
3
x
+
5
équivaut successivement à :
l
(
x
)
=
e
x
−
7
−
2
x
×
e
3
x
+
5
−
(
−
2
x
+
1
)
l\left(x\right)=e^{x-7-2x} \times e^{3x+5-\left(-2x+1\right)}
l
(
x
)
=
e
x
−
7
−
2
x
×
e
3
x
+
5
−
(
−
2
x
+
1
)
l
(
x
)
=
e
x
−
7
−
2
x
×
e
3
x
+
5
+
2
x
−
1
l\left(x\right)=e^{x-7-2x} \times e^{3x+5+2x-1}
l
(
x
)
=
e
x
−
7
−
2
x
×
e
3
x
+
5
+
2
x
−
1
l
(
x
)
=
e
−
x
−
7
×
e
5
x
+
4
l\left(x\right)=e^{-x-7} \times e^{5x+4}
l
(
x
)
=
e
−
x
−
7
×
e
5
x
+
4
l
(
x
)
=
e
−
x
−
7
+
5
x
+
4
l\left(x\right)=e^{-x-7+5x+4}
l
(
x
)
=
e
−
x
−
7
+
5
x
+
4
l
(
x
)
=
e
4
x
−
3
l\left(x\right)=e^{4x-3}
l
(
x
)
=
e
4
x
−
3