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3
2014 MMATHS Tiebreaker p3 - f(\sqrt{x_1x_2}) =\sqrt{f(x_1)f(x_2)}
2014 MMATHS Tiebreaker p3 - f(\sqrt{x_1x_2}) =\sqrt{f(x_1)f(x_2)}
Source:
October 8, 2023
algebra
MMATHS
functional equation
Problem Statement
Let
f
:
R
+
→
R
+
f : R^+ \to R^+
f
:
R
+
→
R
+
be a function satisfying
f
(
x
1
x
2
)
=
f
(
x
1
)
f
(
x
2
)
f(\sqrt{x_1x_2}) =\sqrt{f(x_1)f(x_2)}
f
(
x
1
x
2
)
=
f
(
x
1
)
f
(
x
2
)
for all positive real numbers
x
1
,
x
2
x_1, x_2
x
1
,
x
2
. Show that
f
(
x
1
x
2
.
.
.
x
n
n
)
=
f
(
x
1
)
f
(
x
2
)
.
.
.
f
(
x
n
)
n
f( \sqrt[n]{x_1x_2... x_n}) = \sqrt[n]{f(x_1)f(x_2) ... f(x_n)}
f
(
n
x
1
x
2
...
x
n
)
=
n
f
(
x
1
)
f
(
x
2
)
...
f
(
x
n
)
for all positive integers
n
n
n
and positive real numbers
x
1
,
x
2
,
.
.
.
,
x
n
x_1, x_2,..., x_n
x
1
,
x
2
,
...
,
x
n
.
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