We are solving the equation:

$18000 = 6000 \left(1 + \frac{0.05}{365}\right)^{365t}$

Step-by-Step Solution:

1. Divide both sides by 6000: $\frac{18000}{6000} = \left(1 + \frac{0.05}{365}\right)^{365t}$ $3 = \left(1 + \frac{0.05}{365}\right)^{365t}$

2. Simplify the base of the exponent: $1 + \frac{0.05}{365} \approx 1.000136986$ So the equation becomes: $3 = (1.000136986)^{365t}$

3. Take the natural logarithm ($\ln$) of both sides: $\ln(3) = \ln\left((1.000136986)^{365t}\right)$ Using the property of logarithms $\ln(a^b) = b \ln(a)$, rewrite it as: $\ln(3) = 365t \ln(1.000136986)$

4. Solve for $t$: $t = \frac{\ln(3)}{365 \ln(1.000136986)}$

5. Calculate the values:

   • $\ln(3) \approx 1.0986$
   • $\ln(1.000136986) \approx 0.000136963$
   • $365 \ln(1.000136986) \approx 0.050011$

   Substituting these: $t = \frac{1.0986}{0.050011} \approx 21.96$

Final Answer:

$t \approx 21.96 \, \text{years}$

Let me know if you'd like additional clarification or more steps! 😊

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Related Questions:

1. What is the meaning of the variables in this equation?
2. How does compound interest differ when the compounding period is increased?
3. How would the solution change if the interest rate were higher or lower?
4. Can you solve a similar equation where the final amount and initial investment are different?
5. What are logarithmic properties and why are they used here?

Tip: When solving for time in compound interest, always check if the base of the exponential term simplifies for faster computation!