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2022 AIME Problems
4
L O G A R I T H M S
L O G A R I T H M S
Source:
February 17, 2022
AIME II
Problem Statement
There is a positive real number
x
x
x
not equal to either
1
20
\tfrac{1}{20}
20
1
or
1
2
\tfrac{1}{2}
2
1
such that
log
20
x
(
22
x
)
=
log
2
x
(
202
x
)
.
\log_{20x} (22x)=\log_{2x} (202x).
lo
g
20
x
(
22
x
)
=
lo
g
2
x
(
202
x
)
.
The value
log
20
x
(
22
x
)
\log_{20x} (22x)
lo
g
20
x
(
22
x
)
can be written as
log
10
(
m
n
)
\log_{10} (\tfrac{m}{n})
lo
g
10
(
n
m
)
, where
m
m
m
and
n
n
n
are relatively prime positive integers. Find
m
+
n
m+n
m
+
n
.
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