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

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll note the first equation:

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

We'll note theĀ 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}.