Find the exact solution of the equation (2^x)^2 + (2^x) −12 = 0, and explain why there is only one solution?

Solution for (2^x)^2 + (2^x) –12 =0 is given below:

Assume p =2^x, then

 p^2 + p -12 =0

=>     (p+4)(p-3)=0  (factorizing the quadratic expression p^2 + p -12)                      

=>      p+4=0 or p-3=0

=>     p = -4 or p =3

Substituting2^x for p, we have

      p=2^x,

For p = -4

=>    -4 =2^x

LOG(-4,2) = x, but log. is only define for the range [0, +∞]

Note: LOG(n,b) means log base b of n.

=>     x= LOG(-4,2) is not a solution since log. is only define for the range [0, +∞]

For p =3,

=>    3 =2^x

=>     LOG(3,2) = x

Therefore, the only solution to the given equation is

x= LOG(3,2), (i.e log base2 of3).

 

 

 

we take first y=2^xthe new equation become: y^2 + y -12=0      a=1 ; b=1 ; c=-12delta=b^2 -4* a*c =1+4*12=49 y1=-4   ;   y2=3we return to the x valuenow we have2^x=-4 or3than we put logarithm in the front of each of them -4<0 wich not includ in the domain of defenition of logarithmit remain3>0 than accepted2^x=3 => x*log(2)=3 => x=3/log(2) is the only solutionRq: the domain of defenition of logarithm is (0;+infinity)    

2,-2,