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You need to convert the given exponential equation a quadratic equation, such that:
`5^x = y => 5^(4x) = (5^x)^4 => (5^x)^4 = y^4`
Changing the variable yields:
`y^4 - 2y^2 + 1 = 0 => (y^2 - 1)^2 = 0 => y^2 - 1 = 0 => y^2 = 1 => y_(1,2) = +-1`
Substituting back `y` for `5^x` yields:
`5^x = y_1 => 5^x = 1 => ln 5^x = ln 1 => x*ln 5 = 0 => x = 0`
`5^x = y_2 => 5^x = -1` invalid
Hence, evaluating the solution to exponential equation yields `x = 0.`
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:
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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