What are the roots of the equation 16^x-3*4^x+2=0?

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have the equation 16^x - 3*4^x + 2 = 0 to solve for x.

Now 16 = 4^2

16^x - 3*4^x + 2 = 0

=> 4^2^x - 3*4^x + 2 = 0

let y = 4^x

=> y^2 - 3y + 2 = 0

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

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

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

y is equal to 1 and 2

=> 4^x = 1 and 4^x = 2

=> x = 0 and x = 1/2

Therefore we have x = 0 and x = 1/2.

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find the roots of 16^x-3*4^x+2=0.

We know that 16^x =  (4x)^2.

So we put 4^x = t in the given equation:

t^2-3x+2 = 0.

(t-2)(t-1) = 0.

t-2 = 0, t-1 = 0.

t=2, or t= 1.

t= 2 gives 4^2 = 2= 4^(1/2. So x= 1/2.

t= 1 gives. 4^x= 1 = 4^0. So x= 0.

So x= 0, x= 1/2 are the solutions.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll write 16 = 4^2.

We'll re-write the equation in this manner:

(4^2)^x - 3*4^x + 2 = 0

We'll substitute 4^x by the variable, t.

t^2 - 3t + 2 = 0

We'll apply quadratic formula for finding t:

t1 = [3+sqrt (9-4*2)]/2*1

t1 = [3+sqrt (1)]/2

t1 = (3+1)/2

t1 = 2

t2 = (3-1)/2

t2 = 1

We'll find the values of x, now.

4^x = t1

4^x=2

We'll write 4^x = 2^2x

2^2x = 2^1

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

2x = 1

x = 1/2

4^x = t2

4^x = 1

4^x = 4^0

x = 0

The solutions of the equation are {0 ; 1/2}.

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