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Logarithm/Base/Rules/Fact/Proof/Exercise
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Logarithm/Base/Rules/Fact
Prove that for the logarithm to base
b
{\displaystyle {}b}
the following calculation rules hold.
We have
log
b
(
b
x
)
=
x
{\displaystyle {}\log _{b}(b^{x})=x}
and
b
log
b
(
y
)
=
y
{\displaystyle {}b^{\log _{b}(y)}=y}
, ie, the logarithm to base
b
{\displaystyle {}b}
is the inverse to the exponential function to the base
b
{\displaystyle {}b}
.
We have
log
b
(
y
⋅
z
)
=
log
b
y
+
log
b
z
{\displaystyle {}\log _{b}(y\cdot z)=\log _{b}y+\log _{b}z}
.
We have
log
b
y
u
=
u
⋅
log
b
y
{\displaystyle {}\log _{b}y^{u}=u\cdot \log _{b}y}
for
u
∈
R
{\displaystyle {}u\in \mathbb {R} }
.
We have
log
a
y
=
log
a
(
b
log
b
y
)
=
log
b
y
⋅
log
a
b
.
{\displaystyle {}\log _{a}y=\log _{a}(b^{\log _{b}y})=\log _{b}y\cdot \log _{a}b\,.}
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