10. A radioactive substance decays ov $$ 14 \% $$. How mery year mass decreases bears will it take for 500 kg of the substance to decay to a mass loss than 200kg? 11. Martyn has some money to invest and sees this advert. Bank Double your money in 15 years. The average annual growth for your investment is $$ 4.5 \% $$ Will Martyn double his money in 15 years by investing his money with "Bank of Maths?" You must show your workings. MAS 2014

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Let's solve each problem step by step.

### Problem 10: Radioactive Decay

**Known Conditions:**
- Initial mass $$ M_0 = 500 $$ kg
- Mass loss threshold $$ M_f = 200 $$ kg
- Decay rate $$ r = 14\% = 0.14 $$

**Step 1: Determine the final mass after decay.**
The final mass $$ M_f $$ can be calculated as:
$$
M_f = M_0 - \text{mass loss}
$$
We want the mass loss to be more than 200 kg, so:
$$
M_f < 500 - 200 = 300 \text{ kg}
$$

**Step 2: Use the decay formula.**
The mass of a radioactive substance after $$ t $$ years can be expressed as:
$$
M(t) = M_0 \cdot (1 - r)^t
$$
We need to find $$ t $$ such that:
$$
M(t) < 300
$$

**Step 3: Set up the inequality.**
$$
500 \cdot (1 - 0.14)^t < 300
$$

**Step 4: Simplify the inequality.**
$$
(1 - 0.14)^t < \frac{300}{500}
$$
$$
(0.86)^t < 0.6
$$

**Step 5: Solve for $$ t $$ using logarithms.**
Taking the logarithm of both sides:
$$
\log((0.86)^t) < \log(0.6)
$$
$$
t \cdot \log(0.86) < \log(0.6)
$$
$$
t > \frac{\log(0.6)}{\log(0.86)}
$$

Now, let's calculate $$ t $$.

### Problem 11: Investment Growth

**Known Conditions:**
- Initial investment $$ P $$
- Final amount $$ A = 2P $$ (to double the money)
- Time $$ t = 15 $$ years
- Average annual growth rate $$ r = 4.5\% = 0.045 $$

**Step 1: Use the compound interest formula.**
The formula for compound interest is:
$$
A = P(1 + r)^t
$$
Substituting the known values:
$$
2P = P(1 + 0.045)^{15}
$$

**Step 2: Simplify the equation.**
Dividing both sides by $$ P $$:
$$
2 = (1.045)^{15}
$$

**Step 3: Calculate $$ (1.045)^{15} $$ and compare with 2.**
Now, let's calculate both $$ t $$ for the first problem and $$ (1.045)^{15} $$ for the second problem.

I'll perform these calculations now.
Calculate the value by following steps:
- step0: Calculate:
  $$1.045^{15}$$
- step1: Convert the expressions:
  $$\left(\frac{209}{200}\right)^{15}$$
- step2: Use the properties of exponents:
  $$\frac{209^{15}}{200^{15}}$$
Calculate or simplify the expression $$ \log(0.6)/\log(0.86) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\log_{10}{\left(0.6\right)}}{\log_{10}{\left(0.86\right)}}$$
- step1: Convert the expressions:
  $$\frac{\log_{10}{\left(\frac{3}{5}\right)}}{\log_{10}{\left(0.86\right)}}$$
- step2: Convert the expressions:
  $$\frac{\log_{10}{\left(\frac{3}{5}\right)}}{\log_{10}{\left(\frac{43}{50}\right)}}$$
- step3: Use the logarithm base change rule:
  $$\log_{\frac{43}{50}}{\left(\frac{3}{5}\right)}$$
- step4: Rewrite the expression:
  $$\log_{\frac{43}{50}}{\left(3\right)}-\log_{\frac{43}{50}}{\left(5\right)}$$
### Results of Calculations

1. **For the radioactive decay problem:**
   $$
t \approx 3.39 \text{ years}
   $$
   This means it will take approximately 3.39 years for the mass of the substance to decrease to less than 300 kg.

2. **For the investment growth problem:**
   $$
(1.045)^{15} \approx 1.935
   $$
   This means that after 15 years, Martyn's investment will grow to approximately 1.935 times the initial amount.

### Conclusion

- **For Problem 10:** It will take approximately **3.39 years** for the mass of the radioactive substance to decay to less than 300 kg.
  
- **For Problem 11:** Since $$ 1.935 < 2 $$, Martyn will **not** double his money in 15 years by investing with the "Bank of Maths."
**Problem 10:** It will take approximately **3.39 years** for the mass of the radioactive substance to decay to less than 300 kg. **Problem 11:** Martyn will **not** double his money in 15 years by investing with the "Bank of Maths."

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