Solve the following problems involving exponents and powers.

a) To multiply terms with the same base, add their exponents.

$$a^x \times a^x = a^{x+x} = a^{2x}$$

The answer is $\boxed{a^{2x}}$.

b) Calculate the values of each power and then multiply.

$$5^3 \times 3^2 = (5 \times 5 \times 5) \times (3 \times 3)$$ $$5^3 \times 3^2 = 125 \times 9$$ $$5^3 \times 3^2 = 1125$$

The answer is $\boxed{1125}$.

c) To multiply terms with the same base, add their exponents.

$$c^x \times c^x = c^{x+x} = c^{2x}$$

Given $c^{2x} = c^4$, the exponents must be equal.

$$2x = 4$$ $$x = \frac{4}{2}$$ $$x = 2$$

The answer is $\boxed{x=2}$.

d) Calculate the values of each power and then subtract.

$$2^3 - 2^2 = (2 \times 2 \times 2) - (2 \times 2)$$ $$2^3 - 2^2 = 8 - 4$$ $$2^3 - 2^2 = 4$$

The answer is $\boxed{4}$.

e) To solve for $x$, express 100 as a power of 10.

$$10^x = 100$$ $$10^x = 10^2$$

Since the bases are the same, the exponents must be equal.

$$x = 2$$

The answer is $\boxed{x=2}$.

f) The equation is $6^x - 6^x = 216$.

$$0 = 216$$

This statement is false, so there is no solution for $x$.
The answer is $\boxed{\text{No solution}}$.

g) To solve for $x$, express 16 as a power of 4.

$$4^x = 16$$ $$4^x = 4^2$$

Since the bases are the same, the exponents must be equal.

$$x = 2$$

The answer is $\boxed{x=2}$.

a) Decrease 1000 by 10% then by 20%.
First decrease by 10%:

$$1000 \times (1 - 0.10) = 1000 \times 0.90 = 900$$

Then decrease the new amount (900) by 20%:

$$900 \times (1 - 0.20) = 900 \times 0.80 = 720$$

The answer is $\boxed{720}$.

b) After increasing by 15%, it became 34500. What is the original amount?
Let the original amount be $A$.

$$A \times (1 + 0.15) = 34500$$ $$A \times 1.15 = 34500$$ $$A = \frac{34500}{1.15}$$ $$A = 30000$$

The answer is $\boxed{30000}$.

c) What amount when decreased by 12% becomes 528000?
Let the original amount be $A$.

$$A \times (1 - 0.12) = 528000$$ $$A \times 0.88 = 528000$$ $$A = \frac{528000}{0.88}$$ $$A = 600000$$

The answer is $\boxed{600000}$.

d) When 2000 is decreased by P%, it becomes 16000. Find the value of P.
Assumption: The question likely meant "increased by P%" or the initial value was 20000, as a decrease from 2000 cannot result in 16000. I will proceed assuming it meant "When 2000 is increased by P%, it becomes 16000."
Let the original amount be 2000.

$$2000 \times (1 + \frac{P}{100}) = 16000$$ $$1 + \frac{P}{100} = \frac{16000}{2000}$$ $$1 + \frac{P}{100} = 8$$ $$\frac{P}{100} = 8 - 1$$ $$\frac{P}{100} = 7$$ $$P = 7 \times 100$$ $$P = 700$$

The answer is $\boxed{P=700}$.

e) Increase 48000 by 12.5% then by 2.25%.
First increase by 12.5%:

$$48000 \times (1 + 0.125) = 48000 \times 1.125 = 54000$$

Then increase the new amount (54000) by 2.25%:

$$54000 \times (1 + 0.0225) = 54000 \times 1.0225 = 55215$$

The answer is $\boxed{55215}$.

a) Jin covered a distance of 50 km in 2 hours. Calculate speed in km/hr.
Speed is calculated as distance divided by time.

$$\text{Speed} = \frac{\text{Distance}}{\text{Time}}$$ $$\text{Speed} = \frac{50 \text{ km}}{2 \text{ hr}}$$ $$\text{Speed} = 25 \text{ km/hr}$$

The speed in km/hr is $\boxed{25 \text{ km/hr}}$.

b) What was the speed in m/sec?
Convert 25 km/hr to m/sec.

$$1 \text{ km} = 1000 \text{ m}$$ $$1 \text{ hr} = 3600 \text{ sec}$$ $$\text{Speed} = 25 \frac{\text{km}}{\text{hr}} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ hr}}{3600 \text{ sec}}$$ $$\text{Speed} = \frac{25 \times 1000}{3600} \text{ m/sec}$$ $$\text{Speed} = \frac{25000}{3600} \text{ m/sec}$$ $$\text{Speed} = \frac{250}{36} \text{ m/sec}$$ $$\text{Speed} = \frac{125}{18} \text{ m/sec}$$ $$\text{Speed} \approx 6.94 \text{ m/sec}$$

The speed in m/sec is $\boxed{\frac{125}{18} \text{ m/sec}}$ or approximately $\boxed{6.94 \text{ m/sec}}$.

A cyclist covered 200 m in 10 seconds. Find the constant speed in km/hr.
First, calculate the speed in m/sec.

$$\text{Speed} = \frac{\text{Distance}}{\text{Time}}$$ $$\text{Speed} = \frac{200 \text{ m}}{10 \text{ sec}}$$ $$\text{Speed} = 20 \text{ m/sec}$$

Now, convert 20 m/sec to km/hr.

$$1 \text{ m} = \frac{1}{1000} \text{ km}$$ $$1 \text{ sec} = \frac{1}{3600} \text{ hr}$$ $$\text{Speed} = 20 \frac{\text{m}}{\text{sec}} \times \frac{1 \text{ km}}{1000 \text{ m}} \times \frac{3600 \text{ sec}}{1 \text{ hr}}$$ $$\text{Speed} = \frac{20 \times 3600}{1000} \text{ km/hr}$$ $$\text{Speed} = \frac{72000}{1000} \text{ km/hr}$$ $$\text{Speed} = 72 \text{ km/hr}$$

The constant speed in km/hr is $\boxed{72 \text{ km/hr}}$.