Saturday, November 20, 2010

Verify if the equation has solution? 5^4x-2*25^x+1=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:


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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