Expert Answers
hala718 eNotes educator| Certified Educator

4^x -2^x =56

4^x -2^x -56= 0

Now let us rewrite 4^x as (2^x)^2

Now assume that 2^x = y

==> y^2 -y -56 =0

==> (y-8)(y+7)

==> y1= 8 ==> y1= 2^x1=8 ==> X1= 3

==> Y2= -7 ==> Y2=2^X2=-7 (IMPOSSIBLE)

Then the solution is x= 3

giorgiana1976 | Student

This is an exponential equation and we'll solve it using the substitution technique, but, for the beginning, we'll write 4^x = 2^2x.

2^2x - 2^x = 56

We'll substitute 2^x by t:

t^2 - t - 56 = 0

We'll apply the quadratic formula:

t1 = [1+sqrt(1+224)]/2

t1 = (1+15)/2

t1 = 8

t2 = (1-15)/2

t2 = -7

But 2^x = t

So, 2^x = t1

2^x = 8

2^x = 2^3

We'll use one to one property:

x = 3

2^x = t2

2^x = -7 is undefined because 2^x>0.

So, the equation has only one solution, which is x=3. 

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