38. A savings account earning compound interest triples in value in 10 years. How long will it take for the original investment to quadruple?

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Let the original principal be $$ P $$ and the rate be $$ r $$. The value of the investment after $$ t $$ years is given by  
$$
P \cdot (1+r)^t.
$$

1. Since the investment triples in 10 years, we have  
   $$
   P \cdot (1+r)^{10} = 3P.
   $$
   Dividing both sides by $$ P $$ gives  
   $$
   (1+r)^{10} = 3.
   $$

2. To find the time $$ t $$ for the investment to quadruple, set  
   $$
   P \cdot (1+r)^t = 4P.
   $$
   Dividing both sides by $$ P $$ yields  
   $$
   (1+r)^t = 4.
   $$

3. Express $$ t $$ in terms of logarithms. Taking the logarithm of both sides, we get  
   $$
   \ln((1+r)^t) = \ln(4).
   $$
   Using the logarithm power rule, this becomes  
   $$
   t \ln(1+r) = \ln(4).
   $$
   Therefore,  
   $$
   t = \frac{\ln(4)}{\ln(1+r)}.
   $$

4. Substitute the value of $$ \ln(1+r) $$ found from the tripling condition. From  
   $$
   (1+r)^{10} = 3,
   $$
   taking logs gives  
   $$
   10\ln(1+r) = \ln(3) \quad \Rightarrow \quad \ln(1+r) = \frac{\ln(3)}{10}.
   $$
   Substituting into the expression for $$ t $$, we obtain  
   $$
   t = \frac{\ln(4)}{\frac{\ln(3)}{10}} = \frac{10\ln(4)}{\ln(3)}.
   $$

5. Hence, the time required for the original investment to quadruple is  
   $$
   t = \frac{10\ln(4)}{\ln(3)} \text{ years}.
   $$

6. Numerically, using approximate values $$ \ln(4) \approx 1.3863 $$ and $$ \ln(3) \approx 1.0986 $$, we have  
   $$
   t \approx \frac{10 \times 1.3863}{1.0986} \approx 12.61 \text{ years}.
   $$

Thus, it will take approximately $$ 12.61 $$ years for the original investment to quadruple.
It will take approximately 12.61 years for the original investment to quadruple.

38. A savings account earning compound interest triples in value in 10 years. How long will it take for the original investment to quadruple?

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