The given exponential equation requires
substitution technique to compute it's
roots.
Now, we notice that 25 =
5^2
We'll re-write the equation
as:
5^4x - 2*5^2x + 1 = 0
It
is a bi-quadratic equation:
We'll substitute 5^2x by
another variable.
5^2x =
t
We'll square raise both
sides:
5^4x =t^2
We'll
re-write the equation, having "t" as variable.
t^2 - 2t + 1
= 0
The equation above is the result of expanding the
square:
(t-1)^2 = 0
t1 = t2 =
1
But 5^2x = t1.
5^2x =
1
We'll write 1 as a power of
5:
5^2x = 5^0
Since the bases
are matching, we'll apply the one to one property:
2x =
0
We'll divide by 2:
x =
0.
The equation has a solution and it is x =
0.
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