2^x - 79 = -4^x

==> 4^x + 2^x - 79 = 0

Let y= e^x ==> 4^x = y^2

==> y^2 + 2y - 79 = 0

==> y1= [-2+ sqrt(4+316)]/2 = -1 + sqrt(320)/2 = 7.88 = 8 (approx.) ==> 8 = 2^x1 ==> x1= 3

==> y2= -1 - 8.88 = -10 (approx.) ==> -10 = 2^x2 (impossible)

Then the only answer is:

x = 3 (approx.)

This is an exponential equation and we'll re-write it, moving all terms to one side:

4^x + 2^x - 79 = 0

We'll note 2^x=t => 4^x=(2^2)^x=(2^x)^2=t^2

We'll write the equivalent expression:

t^2 + t + 1 = 80

t^2 + t + 1 - 80 = 0

t^2 + t - 79 = 0

We'll apply the quadratic formula:

t1 = [-1+sqrt(1+1+316)]/2=16/2=approx. 8

2^x=t1

2^x=8 => 2^x=2^3 =>x=3

t2=(-1-17)/2=approx. -9

But 2^x>0 so it's impossible that 2^x=-9!

The equation will have the solution x=3 (approx.)

2^x-79 = -4^x.

To solve for x.

Rewrite the given equation as:

2^x+4^x -79 = 0..........(1)

4^x = (2^x)^2. Put 2^x in eq (1):

t+t^2-79 =0

t^2+t-79 = 0

t = {-1+or- sqrt(1+4*1*79)}/2 = (-1+or - sqrt317)/2

t = (-1+sqrt 317)/2 Or t = (-1-sqrt317)/2.

Therefore,

2^x = (-1+sqrt317)/2 =

x log2 = log {[-1+sqrt317)/2]

x = log {[-1+sqrt317)/2] / log2