urgent help!!!!!!!!?

Question

solve for s and p given that: 5^p*25^s = 1 and 16^p/2^s = 2

Answer 1

5^p * 25^s = 1

5^p * 5^(2*s) = 1

5^p * 5^(2*s) = 1

5^(p+2*s) = 1

p + 2*s = 0

---

16^p/2^s = 2

2^(4*p)/2^s = 2

2^(4*p-s) = 2

4*p-s = 1

---

Now you have to solve the system:

p + 2*s = 0
4*p-s = 1


p = -2*s

4*(-2*s) - s = 1

-8s-s=1
-9s=1

s= -1/9

p = -2*s = 2/9

Answer 2

16^p/2^s = 2
16^p = 2^s * 2
2^(4p) = 2^(s + 1)

4p = s + 1 ..... (1)

5^p * 25^s = 1
5^p * 5^(2s) = 1
5^(p + 2s) = 1

Since anything to the power 0 is 1, and base is 5, it is true that:

p + 2s = 0 ..... (2)

We must solve (1) and (2)

(2) * 4,
4p + 8s = 0.....(3)

Put (1) in (3)
s + 1 + 8s = 0
9s + 1 = 0
s = -1/9

Put s = -1/9 in (1)
4p = 1 - 1/9
4p = 8/9
p = 2/9

s = -1/9, p = 1/9

Answer 3

5^p*25^s = 1
5^p*5^2s=1
5^(p+2s)=1
5^(p+2s)=5^0
p+2s=0.... eqn 1

16^p/2^s = 2
2^4p/2^s=2
2^(4p-s)=2
2^(4p-s)=2^1
4p-s=1 ...... eqn 2

eqn 1 &2 are simultaneous Equations

fom eqn 2,
s=4p-1

replacing s in eqn 1
p+2(4p-1)=0
p+8p - 2 = 0
9p = 2
p=2/9

from eqn 2
s=4p-1= 8/9 - 1 = -1/9

so, p=2/9 and s=(-1/9)

Answer 4

I am not sure this is correct. See the other answerer's work.

5^p x 25^s = 1

16^p/2^s = 2
________________

5^p x 25 ^s = 1
125^ p + s = 1
To make this equation true,
p + s = 0
_______________
16^p/2^s = 2
8^p - s = 2
To make this equation true,
p -s = 1/3
______________

p + s = 0 (eq. 1)
p - s = 1/3 (eq. 2)

Add eq. 1 and eq. 2

p + s = 0
+ p - s = 1/3
_____________
2p = 1/3
p = (1/3)/2
p = 1/3 x 1/2
p = 1/6

p + s = 0
1/6 + s = 0
s = - 1/6
_____________
p = 1/6
s = -1/6

Answer 5

log(5^p * 25^s) = log(1)
plog(5) + s(log(25) = log(1)
plog(5) + 2slog(5) = log(1)
(p + 2s)log(5) = log(1)
log(1) = 0 ,
p + 2s = 0
p = -2s or s = -p/2

log(16^p / 2^s) = log(2)
SImilar to the previous one, you end up with:
4plog(2) - slog(2) = log(2)
4p - s = 1
s = 4p -1 or p = (s + 1)/4

If these are both true:

s = 4p -1 = 4(-2s) -1 = -8s - 1
9s = -1
s = -1/9
p = -2s = 2/9