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Problem
Problem
If
x
1
x_1
x
1
and
x
2
x_2
x
2
(
x
2
>
x
1
x_2 > x_1
x
2
>
x
1
) are the integral solutions of the equation
(
log
5
x
)
2
+
log
5
(
5
x
)
=
1
,
the value of
∣
x
2
−
4
x
1
∣
is
(\log_5 x)^2 + \log_5\left(\frac{5}{x}\right) = 1,\ \text{the value of } |x_2 - 4x_1| \text{ is}
(
lo
g
5
x
)
2
+
lo
g
5
(
x
5
)
=
1
,
the value of
∣
x
2
−
4
x
1
∣
is
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