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

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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}.**

To solve the equations:

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

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

We put 5^x= u and 7^y = v. Then 25^x = (5^x)^2 = u^2 and 40^x = (7^y)^2 = v^2.

There with the above transformations, the given equations become:

u+v = 4 and 2(u^2+v^2) = 20. Or u^2+v^2 = 20/2 = 10.

Therefore (u+v)^2 = u^2+v^2 +2uv.

So 2uv = (u+v)^2 - (u^2+v^2).

2uv = 4^2 - 10 = 16-10 = 6.

Therefore (u-v)^2 = (u^2+v^2)-2uv = 10-6 = 4.

Therefore u-v = sqrt(u-v)^2 = sqrt4 = 2.

So u+v = 4 and u-v = 2. Adding these, 2u =4+2 = 6. Or u = 6/2 = 3.

(u+v)-(u-v) = 2v = 4-2 = 2. So v = 2/1 = 1.

Therefore u = 5^x = 2. So xlog5 = log3 or x = log3/log5 = ln3/ln5.

v = 7^y = 1 = 7^0. So y = o.

Therefore x = log3/log5 or ln3/ln5 and y = 0.