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Fonction logarithme népérien
Propriétés algébriques - Exercice 1
25 min
35
Simplifier les expressions suivantes :
Question 1
a
(
x
)
=
ln
(
2
)
+
ln
(
5
)
a\left(x\right)=\ln \left(2\right)+\ln \left(5\right)
a
(
x
)
=
ln
(
2
)
+
ln
(
5
)
Correction
ln
(
a
)
+
ln
(
b
)
=
ln
(
a
×
b
)
\ln \left(a\right)+\ln \left(b\right)=\ln \left(a\times b\right)
ln
(
a
)
+
ln
(
b
)
=
ln
(
a
×
b
)
ln
(
a
)
−
ln
(
b
)
=
ln
(
a
b
)
\ln \left(a\right)-\ln \left(b\right)=\ln \left(\frac{a}{b} \right)
ln
(
a
)
−
ln
(
b
)
=
ln
(
b
a
)
ln
(
1
a
)
=
−
ln
(
a
)
\ln \left(\frac{1}{a} \right)=-\ln \left(a\right)
ln
(
a
1
)
=
−
ln
(
a
)
ln
(
a
n
)
=
n
ln
(
a
)
\ln \left(a^{n} \right)=n\ln \left(a\right)
ln
(
a
n
)
=
n
ln
(
a
)
1
2
ln
(
a
)
=
ln
(
a
)
\frac{1}{2} \ln \left(a\right)=\ln \left(\sqrt{a} \right)
2
1
ln
(
a
)
=
ln
(
a
)
e
ln
a
=
a
e^{\ln a} =a
e
l
n
a
=
a
a
(
x
)
=
ln
(
2
×
5
)
⇔
a\left(x\right)=\ln \left(2\times 5\right)\Leftrightarrow
a
(
x
)
=
ln
(
2
×
5
)
⇔
a
(
x
)
=
ln
(
10
)
a\left(x\right)=\ln \left(10\right)
a
(
x
)
=
ln
(
10
)
Question 2
b
(
x
)
=
ln
(
3
)
−
ln
(
4
)
b\left(x\right)=\ln \left(3\right)-\ln \left(4\right)
b
(
x
)
=
ln
(
3
)
−
ln
(
4
)
Correction
ln
(
a
)
+
ln
(
b
)
=
ln
(
a
×
b
)
\ln \left(a\right)+\ln \left(b\right)=\ln \left(a\times b\right)
ln
(
a
)
+
ln
(
b
)
=
ln
(
a
×
b
)
ln
(
a
)
−
ln
(
b
)
=
ln
(
a
b
)
\ln \left(a\right)-\ln \left(b\right)=\ln \left(\frac{a}{b} \right)
ln
(
a
)
−
ln
(
b
)
=
ln
(
b
a
)
ln
(
1
a
)
=
−
ln
(
a
)
\ln \left(\frac{1}{a} \right)=-\ln \left(a\right)
ln
(
a
1
)
=
−
ln
(
a
)
ln
(
a
n
)
=
n
ln
(
a
)
\ln \left(a^{n} \right)=n\ln \left(a\right)
ln
(
a
n
)
=
n
ln
(
a
)
1
2
ln
(
a
)
=
ln
(
a
)
\frac{1}{2} \ln \left(a\right)=\ln \left(\sqrt{a} \right)
2
1
ln
(
a
)
=
ln
(
a
)
e
ln
a
=
a
e^{\ln a} =a
e
l
n
a
=
a
b
(
x
)
=
ln
(
3
4
)
b\left(x\right)=\ln \left(\frac{3}{4} \right)
b
(
x
)
=
ln
(
4
3
)
Question 3
c
(
x
)
=
3
ln
(
2
)
+
2
ln
(
3
)
−
1
2
ln
(
9
)
c\left(x\right)=3\ln \left(2\right)+2\ln \left(3\right)-\frac{1}{2} \ln \left(9\right)
c
(
x
)
=
3
ln
(
2
)
+
2
ln
(
3
)
−
2
1
ln
(
9
)
Correction
ln
(
a
)
+
ln
(
b
)
=
ln
(
a
×
b
)
\ln \left(a\right)+\ln \left(b\right)=\ln \left(a\times b\right)
ln
(
a
)
+
ln
(
b
)
=
ln
(
a
×
b
)
ln
(
a
)
−
ln
(
b
)
=
ln
(
a
b
)
\ln \left(a\right)-\ln \left(b\right)=\ln \left(\frac{a}{b} \right)
ln
(
a
)
−
ln
(
b
)
=
ln
(
b
a
)
ln
(
1
a
)
=
−
ln
(
a
)
\ln \left(\frac{1}{a} \right)=-\ln \left(a\right)
ln
(
a
1
)
=
−
ln
(
a
)
ln
(
a
n
)
=
n
ln
(
a
)
\ln \left(a^{n} \right)=n\ln \left(a\right)
ln
(
a
n
)
=
n
ln
(
a
)
1
2
ln
(
a
)
=
ln
(
a
)
\frac{1}{2} \ln \left(a\right)=\ln \left(\sqrt{a} \right)
2
1
ln
(
a
)
=
ln
(
a
)
e
ln
a
=
a
e^{\ln a} =a
e
l
n
a
=
a
c
(
x
)
=
3
ln
(
2
)
+
2
ln
(
3
)
−
1
2
ln
(
9
)
c\left(x\right)=3\ln \left(2\right)+2\ln \left(3\right)-\frac{1}{2} \ln \left(9\right)
c
(
x
)
=
3
ln
(
2
)
+
2
ln
(
3
)
−
2
1
ln
(
9
)
équivaut successivement à
c
(
x
)
=
ln
(
2
3
)
+
ln
(
3
2
)
−
ln
(
9
)
c\left(x\right)=\ln \left(2^{3} \right)+\ln \left(3^{2} \right)-\ln \left(\sqrt{9} \right)
c
(
x
)
=
ln
(
2
3
)
+
ln
(
3
2
)
−
ln
(
9
)
c
(
x
)
=
ln
(
8
)
+
ln
(
9
)
−
ln
(
3
)
c\left(x\right)=\ln \left(8\right)+\ln \left(9\right)-\ln \left(3\right)
c
(
x
)
=
ln
(
8
)
+
ln
(
9
)
−
ln
(
3
)
c
(
x
)
=
ln
(
8
×
9
)
−
ln
(
3
)
c\left(x\right)=\ln \left(8\times9\right)-\ln \left(3\right)
c
(
x
)
=
ln
(
8
×
9
)
−
ln
(
3
)
c
(
x
)
=
ln
(
72
)
−
ln
(
3
)
c\left(x\right)=\ln \left(72\right)-\ln \left(3\right)
c
(
x
)
=
ln
(
72
)
−
ln
(
3
)
c
(
x
)
=
ln
(
72
3
)
c\left(x\right)=\ln \left(\frac{72}{3} \right)
c
(
x
)
=
ln
(
3
72
)
c
(
x
)
=
ln
(
24
)
c\left(x\right)=\ln \left(24\right)
c
(
x
)
=
ln
(
24
)
Question 4
d
(
x
)
=
e
ln
3
−
e
ln
6
d\left(x\right)=e^{\ln 3} -e^{\ln 6}
d
(
x
)
=
e
l
n
3
−
e
l
n
6
Correction
ln
(
a
)
+
ln
(
b
)
=
ln
(
a
×
b
)
\ln \left(a\right)+\ln \left(b\right)=\ln \left(a\times b\right)
ln
(
a
)
+
ln
(
b
)
=
ln
(
a
×
b
)
ln
(
a
)
−
ln
(
b
)
=
ln
(
a
b
)
\ln \left(a\right)-\ln \left(b\right)=\ln \left(\frac{a}{b} \right)
ln
(
a
)
−
ln
(
b
)
=
ln
(
b
a
)
ln
(
1
a
)
=
−
ln
(
a
)
\ln \left(\frac{1}{a} \right)=-\ln \left(a\right)
ln
(
a
1
)
=
−
ln
(
a
)
ln
(
a
n
)
=
n
ln
(
a
)
\ln \left(a^{n} \right)=n\ln \left(a\right)
ln
(
a
n
)
=
n
ln
(
a
)
1
2
ln
(
a
)
=
ln
(
a
)
\frac{1}{2} \ln \left(a\right)=\ln \left(\sqrt{a} \right)
2
1
ln
(
a
)
=
ln
(
a
)
e
ln
a
=
a
e^{\ln a} =a
e
l
n
a
=
a
d
(
x
)
=
e
ln
3
−
e
ln
6
⇔
d
(
x
)
=
3
−
6
⇔
d\left(x\right)=e^{\ln 3} -e^{\ln 6}\Leftrightarrow d\left(x\right)=3-6\Leftrightarrow
d
(
x
)
=
e
l
n
3
−
e
l
n
6
⇔
d
(
x
)
=
3
−
6
⇔
d
(
x
)
=
−
3
d\left(x\right)=-3
d
(
x
)
=
−
3
Question 5
f
(
x
)
=
e
2
+
ln
(
4
)
e
ln
2
f\left(x\right)=\frac{e^{2+\ln \left(4\right)} }{e^{\ln 2} }
f
(
x
)
=
e
l
n
2
e
2
+
l
n
(
4
)
Correction
ln
(
a
)
+
ln
(
b
)
=
ln
(
a
×
b
)
\ln \left(a\right)+\ln \left(b\right)=\ln \left(a\times b\right)
ln
(
a
)
+
ln
(
b
)
=
ln
(
a
×
b
)
ln
(
a
)
−
ln
(
b
)
=
ln
(
a
b
)
\ln \left(a\right)-\ln \left(b\right)=\ln \left(\frac{a}{b} \right)
ln
(
a
)
−
ln
(
b
)
=
ln
(
b
a
)
ln
(
1
a
)
=
−
ln
(
a
)
\ln \left(\frac{1}{a} \right)=-\ln \left(a\right)
ln
(
a
1
)
=
−
ln
(
a
)
ln
(
a
n
)
=
n
ln
(
a
)
\ln \left(a^{n} \right)=n\ln \left(a\right)
ln
(
a
n
)
=
n
ln
(
a
)
1
2
ln
(
a
)
=
ln
(
a
)
\frac{1}{2} \ln \left(a\right)=\ln \left(\sqrt{a} \right)
2
1
ln
(
a
)
=
ln
(
a
)
e
ln
a
=
a
e^{\ln a} =a
e
l
n
a
=
a
Ici on utilise également les règles sur les exponentielles
e
a
+
b
=
e
a
×
e
b
e^{a+b} =e^{a} \times e^{b}
e
a
+
b
=
e
a
×
e
b
f
(
x
)
=
e
2
+
ln
(
4
)
e
ln
2
f\left(x\right)=\frac{e^{2+\ln \left(4\right)} }{e^{\ln 2} }
f
(
x
)
=
e
l
n
2
e
2
+
l
n
(
4
)
équivaut successivement à
f
(
x
)
=
e
2
×
e
ln
(
4
)
2
f\left(x\right)=\frac{e^{2} \times e^{\ln \left(4\right)} }{2}
f
(
x
)
=
2
e
2
×
e
l
n
(
4
)
f
(
x
)
=
e
2
×
4
2
f\left(x\right)=\frac{e^{2} \times 4}{2}
f
(
x
)
=
2
e
2
×
4
f
(
x
)
=
2
e
2
f\left(x\right)=2e^{2}
f
(
x
)
=
2
e
2
Question 6
g
(
x
)
=
e
ln
(
x
+
1
)
e
ln
(
x
)
g\left(x\right)=e^{\ln \left(x+1\right)} e^{\ln \left(x\right)}
g
(
x
)
=
e
l
n
(
x
+
1
)
e
l
n
(
x
)
Correction
ln
(
a
)
+
ln
(
b
)
=
ln
(
a
×
b
)
\ln \left(a\right)+\ln \left(b\right)=\ln \left(a\times b\right)
ln
(
a
)
+
ln
(
b
)
=
ln
(
a
×
b
)
ln
(
a
)
−
ln
(
b
)
=
ln
(
a
b
)
\ln \left(a\right)-\ln \left(b\right)=\ln \left(\frac{a}{b} \right)
ln
(
a
)
−
ln
(
b
)
=
ln
(
b
a
)
ln
(
1
a
)
=
−
ln
(
a
)
\ln \left(\frac{1}{a} \right)=-\ln \left(a\right)
ln
(
a
1
)
=
−
ln
(
a
)
ln
(
a
n
)
=
n
ln
(
a
)
\ln \left(a^{n} \right)=n\ln \left(a\right)
ln
(
a
n
)
=
n
ln
(
a
)
1
2
ln
(
a
)
=
ln
(
a
)
\frac{1}{2} \ln \left(a\right)=\ln \left(\sqrt{a} \right)
2
1
ln
(
a
)
=
ln
(
a
)
e
ln
a
=
a
e^{\ln a} =a
e
l
n
a
=
a
g
(
x
)
=
e
ln
(
x
+
1
)
e
ln
(
x
)
g\left(x\right)=e^{\ln \left(x+1\right)} e^{\ln \left(x\right)}
g
(
x
)
=
e
l
n
(
x
+
1
)
e
l
n
(
x
)
équivaut successivement à
g
(
x
)
=
(
x
+
1
)
×
(
x
)
g\left(x\right)=\left(x+1\right)\times \left(x\right)
g
(
x
)
=
(
x
+
1
)
×
(
x
)
g
(
x
)
=
x
2
+
x
g\left(x\right)=x^{2} +x
g
(
x
)
=
x
2
+
x
Question 7
h
(
x
)
=
ln
(
3
−
5
)
+
ln
(
3
+
5
)
h\left(x\right)=\ln \left(3-\sqrt{5} \right)+\ln \left(3+\sqrt{5} \right)
h
(
x
)
=
ln
(
3
−
5
)
+
ln
(
3
+
5
)
Correction
ln
(
a
)
+
ln
(
b
)
=
ln
(
a
×
b
)
\ln \left(a\right)+\ln \left(b\right)=\ln \left(a\times b\right)
ln
(
a
)
+
ln
(
b
)
=
ln
(
a
×
b
)
ln
(
a
)
−
ln
(
b
)
=
ln
(
a
b
)
\ln \left(a\right)-\ln \left(b\right)=\ln \left(\frac{a}{b} \right)
ln
(
a
)
−
ln
(
b
)
=
ln
(
b
a
)
ln
(
1
a
)
=
−
ln
(
a
)
\ln \left(\frac{1}{a} \right)=-\ln \left(a\right)
ln
(
a
1
)
=
−
ln
(
a
)
ln
(
a
n
)
=
n
ln
(
a
)
\ln \left(a^{n} \right)=n\ln \left(a\right)
ln
(
a
n
)
=
n
ln
(
a
)
1
2
ln
(
a
)
=
ln
(
a
)
\frac{1}{2} \ln \left(a\right)=\ln \left(\sqrt{a} \right)
2
1
ln
(
a
)
=
ln
(
a
)
e
ln
a
=
a
e^{\ln a} =a
e
l
n
a
=
a
h
(
x
)
=
ln
(
3
−
5
)
+
ln
(
3
+
5
)
h\left(x\right)=\ln \left(3-\sqrt{5} \right)+\ln \left(3+\sqrt{5} \right)
h
(
x
)
=
ln
(
3
−
5
)
+
ln
(
3
+
5
)
équivaut successivement à
h
(
x
)
=
ln
(
(
3
−
5
)
×
(
3
+
5
)
)
h\left(x\right)=\ln \left(\left(3-\sqrt{5} \right)\times\left(3+\sqrt{5} \right)\right)
h
(
x
)
=
ln
(
(
3
−
5
)
×
(
3
+
5
)
)
.
Ensuite il faut utiliser l'identité remarquable :
(
a
−
b
)
×
(
a
−
b
)
=
a
2
−
b
2
\left(a-b\right)\times \left(a-b\right)=a^{2}-b^{2}
(
a
−
b
)
×
(
a
−
b
)
=
a
2
−
b
2
h
(
x
)
=
ln
(
3
2
−
(
5
)
2
)
h\left(x\right)=\ln \left(3^{2} -\left(\sqrt{5} \right)^{2} \right)
h
(
x
)
=
ln
(
3
2
−
(
5
)
2
)
h
(
x
)
=
ln
(
9
−
5
)
h\left(x\right)=\ln \left(9-5\right)
h
(
x
)
=
ln
(
9
−
5
)
h
(
x
)
=
ln
(
4
)
h\left(x\right)=\ln \left(4\right)
h
(
x
)
=
ln
(
4
)