Verify if exponential equation has any roots. 4^x=20-4^(3-x)

We'll re-write the term 4^(3-x) recalling the quotient property
of exponentials:

4^(3-x) =
4^3/4^x

We'll re-write the
equation:

4^3/4^x + 4^x - 20 = 0

We'll
multiply by 4^x both sides:

4^3 + 4^2x - 20*4^x =
0

We'll substitute 4^x = t:

t^2 - 20t +
64 = 0

We'll apply quadratic rule:

t1 =
[20+sqrt(400 - 256)]/2

t1 =
(20+sqrt144)/2

t1 = (20+12)/2

t1 =
16

t2 = (20-12)/2

t2 =
4

But t1 = 4^x => 16 =
4^x

We'll put 16 =
4^2

4^x = 4^2

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

x =
2

t2 = 4 => 4^x = 4 => x =
1

The equation has 2 real solutions and they are: {1 ;
2}.