Solve the simultaneous equations x^(x^2-y^2+8x+1)=1, 2^y=8*2^x

We'll note the first
equation:

x^(x^2-y^2+8x+1)=1
(1)

We'll note the 2nd
equation:

2^y=8*2^x (2)

We'll
take natural logarithms both sides of the first
equation:

ln x^(x^2-y^2+8x+1) = ln
1

(x^2-y^2+8x+1)*ln x =
0

We'll put each factor as
zero:

ln x = 0

x =
e^0

x = 1

x^2-y^2+8x+1 =
0

We'll re-write the x^2-y^2 =
(x-y)(x+y)

2^y=8*2^x

We'll
re-write the second equation, using the quotient rule of the exponential
functions:

2^y/2^x = 8

2^(y -
x) = 2^3

y - x = 3

We'll
multiply by -1:

x - y = -3
(3)

x = y - 3 (4)

We'll
substitute (3) and (4) in the following
relation:

(x-y)(x+y) + 8x + 1 =
0

-3(y-3+y) + 8(y - 3) + 1 =
0

-6y + 9 + 8y - 24 + 1 =
0

We'll combine like terms:

2y
- 14 = 0

2y = 14

y =
7

x = 7 - 3

x =
4

y^2 = x^2 + 8x + 1

y^2 = 1 +
8 + 1

y^2 = 10

y^2 = 16 + 32 +
1

y^2 = 49

y =
7

The solutions of the equation are: {x=1 ;
x=4 ; y=7}.