Find solutions of equation? 14^14x - 2*14^7x + 1=0

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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14^14x - 2*14^7x + 1 = 0

We know that:

14^14x = 14^2*7x = (14^7)^2

==> (14^7x)^2 - 2*14^7x +1 = 0

Let y = 14^7x

==> y^2 - 2y + 1 = 0

==> (y-1)^2 = 0

==>  y=1

But y= 14^7x = 1

==> 7x = 0

==> x= 0

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william1941 | College Teacher | (Level 3) Valedictorian

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To solve the given equation : 14^14x - 2*14^7x + 1=0

substitute 14^7x with Y.

We find that 14^14x = 14^2*7x or 14^7x^2 = Y^2

So the equation can now be written as:

14^14x - 2*14^7x + 1=0

=> Y^2 - 2Y + 1 =0

=> (Y - 1)^0 =0

=> Y - 1 =0

=> Y =1

But Y = 14^7x

Therefore 14^7x =1 .

Now we know that if a^b =1 => b= 0

Therefore as 14^7x = 1 => 7x = 0 => x =0.

Therefore the solution for the equation is x=0.

 

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

14^14x-2*14^7x+1 = 0

To solve this, put 14^7x = y.

Then 14^14x  = (14^7x)^2 = y^2.

Therefore the given equation becomes:

y^2-2y +1 = 0.

(y-1)^2 = 0

y -1 = 0

Therefore y =1.

Therefore y = 1 implies 14^7x = 1.

14^7x = 14^0, as 1= a^0 for any a but not a = 0.

Equate the powers:

7x = 0

x = 0/7 = 0

Therefore x = 0

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

This is an exponential equation that requires substitution technique.

14^14x - 2*14^7x + 1 = 0

It is a bi-quadratic equation:

We'll substitute 14^7x  by another variable.

14^7x  = t

We'll square raise both sides:

14^14x  = 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 14^7x = t1.

14^7x = 1

We'll write 1 as a power of 14:

14^7x= 14^0

Since the bases are matching, we'll apply the one to one property:

7x = 0

We'll divide by 7 both sides:

x = 0.

The solution of the equation is x = 0.

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