# Solve the equation 2^x + 4^(x+1)/2 = 12

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To solve

2^x+4^(x+1)/2 = 12.

Rewrite as:

2^x+4*4^x /2 = 12

2^x+2*(2^x)^2 = 12 , as 4^x = (2^2)^x = (2^x)^2

y +2y^2-12 = 0, where y = 2^x.

2y^2 +y-12 = 0

y = {-1+sqrt(1+4*2*12)}/4

y  = {-1+sqrt97}/4

y = {-1-sqrt97}/4

Therefore  2  ^x = (-1+sqrt97)/4

2log2 = {log(-1+sqrt97/4} / log 2

x = log {(-1+sqrt97)/4 }/ log2.

But if we take the middle term, 4^(x+1)/2 = 4^((x+1)/2):

4 = 2^2.  So 4^((x+1)/2) = 2^2((x+1)/2) = 2^(x+1).

Now the equation could be rewritten as:

2^x+2^(x+1) = 12

(2^x)(1+2)  = 12

(2^x)(3) =12

2^x = 12/3 = 4 = 2^2

2^x = 2^2

x = 2

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

This is an exponential equation. We notice that 4 = 2^2.

We'll re-write 4^[(x+1)/2] = 2^2[(x+1)/2]

We'll simplify and we'll get:

4^[(x+1)/2] = 2^(x+1) = 2*2^x

We'll re-write the equation;

2^x + 4^(x+1)/2 = 12

2^x + 2*2^x = 12

We'll factorize by 2^x:

2^x*(1+2) = 12

2^x*3 = 12

We'll divide by 3 both sides:

2^x = 4

We'll write 4 as a power of 2:

2^x = 2^2

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

x = 2

The solution of the exponential equation is x = 2.