# simultaneous equationsSolve the simultaneous equations 5^x + 7^y = 4 2*25^x + 2*49^y = 20

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We have to solve

5^x + 7^y = 4...(1)

2*25^x + 2*49^y = 20 ...(2)

Let 5^x = a and 7^y = b

(1) => a + b = 4 and

(2) => 2*a^2 + 2*b^2 = 20

=> a^2 + b^2 = 10

Substitute a = 4 - b

=> (4 - b)^2 + b^2 = 10

=> 16 - 8b + b^2 + b^2 = 10

=> 2b^2 - 8b + 6 = 0

=> b^2 - 4b + 3 = 0

=> b^2 - 3b - b + 3 = 0

=> b(b - 3) -1(b - 3) = 0

=> (b - 1)(b -3) = 0

=> b = 1 and b = 3

a = 3 and a = 1

5^x = 1 and 7^y = 3

=> x = 0 and y = log 3/log 7

5^x = 3 and 7^y = 1

=> x = log 3/log 5 and y = 0

**The solution of the equations is (0, log 3/log 7) and (log 3/log 5 , 0)**

We'll note the terms from the first equation as:

5^x = u and 7^y = v.

We notice that if we square the terms from the first equation, we'll obtain the terms from the second equation:

(5^2)^x = u^2 and (7^2)^y = v^2

We'll re-write the system using the new variables u and v:

u + v = 4 (1)

2u^2 + 2v^2 = 20

We'll divide by 2 and we'll get:

u^2 + v^2 = 10

But u^2 + v^2 = (u+v)^2 - 2uv

u^2 + v^2 = 16 - 2uv

16 - 2uv = 10

2uv = 16 - 10

uv = 3 (2)

We'll write u with respect to v, from (1):

u = 4 - v (3)

We'll substitute (3) in (2):

(4 - v)*v = 3

We'll remove the brackets:

4v - v^2 - 3 = 0

We'll multiply by -1 and we'll re-arrange the terms:

v^2 - 4v + 3 = 0

We'll apply the quadratic formula:

v1 = [4+sqrt(16 - 12)]/2

v1 = (4+2)/2

v1 = 3

v2 = 1

7^y = v1

7^y = v2

7^y = 3

We'll take logarithms both sides:

ln 7^y = ln 3

y*ln 7= ln 3

y = ln 3/ln 7

7^y = 1

7^y = 7^0

y = 0

Now, we'll calculate u and then, x:

u1 = 4 - v1

u1 = 4 - 3

u1 = 1

u2 = 4 - v2

u2 = 4 - 1

u2 = 3

We'll calculate x:

5^x = 1

x = 0

5^x = 3

We'll take logarithms both sides:

ln 5^x = ln 3

x*ln 5 = ln 3

x = ln 3/ln 5

The solutions of the system are: {0; ln3/ln7} and {ln 3/ln 5;0}.