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What are the solutions of the equation 2*16^x = 4^x + 1 ?

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lillyan | Student, Undergraduate | (Level 2) Honors

Posted November 8, 2010 at 1:27 AM via web

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What are the solutions of the equation 2*16^x = 4^x + 1 ?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted November 8, 2010 at 1:31 AM (Answer #1)

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We notice that 16=4^2!

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

2*(4^2)^x - 4^x - 1=0

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

2*t^2 - t - 1=0

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

t1=[1+sqrt (9)]/4

t1=(1+3)/4

t1=1

t2=[1-sqrt (1+4*2)]/4

t2=(1-3)/4

t2=-1/2

We didn't find the values of x, yet!

4^x=1

4^x=4^0

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

x = 0

4^x=-1/2

The exponential 4^x is always positive, for any value of x, so, we'll reject the second solution.

The equation has just one solution. The only solution is x= 0.

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neela | High School Teacher | (Level 3) Valedictorian

Posted November 8, 2010 at 1:40 AM (Answer #2)

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To solve 2*16^x = 4^x+1.

Solution:

We know that 16^x = (4^2)^x

16^x = (4^x)^2.

So we put 4^x = t in the given equation, 2*16^x = 4^x+1:

2t^2 = t +1.

2t^2-t-1 = 0.

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

We equate each factor to zero.

2t+1 = 0 . t-1 = 0.

2t+1 = 0 gives t = -1/2 gives no real solution.

 t-1 = 0 gives: t= 1. Or 4^x = 1 = 4^0.

4^x = 4^0 implies x = 0.

So x = 0 is the solution.

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sociality | High School Teacher | (Level 1) Valedictorian

Posted November 8, 2010 at 1:31 AM (Answer #3)

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We need to determine x for 2*16^x = 4^x + 1

Now 2*16^x = 4^x + 1

=> 2*4^2^x = 4^x + 1

=> 2* 4^x^2 = 4^x + 1

let Y = 4^x

=> 2Y^2 = Y + 1

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

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

=> 2Y( Y - 1) + 1(Y - 1) =0

=> (2Y + 1)(Y - 1) = 0

Y = -1/2 or Y = 1

For Y = -1/2 we have 4^x = -1/2 which is not possible.

For Y = 1 we have 4^x = 1 which gives x = 0 as any number raised to the power 0 is 1.

Therefore x = 0

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