Joseph is going to invest $$ \$ 890 $$ and leave it in an account for 15 years. Assuming the interest is compounded annually, what interest rate, to the nearest tenth of a percent, would be required in order for Joseph to end up with $$ \$ 1,710 $$ ?

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1. Write down the compound interest formula:
   $$
   A = P(1 + r)^t
   $$
   where:
   - $$A = 1710$$ is the final amount,
   - $$P = 890$$ is the initial investment,
   - $$r$$ is the annual interest rate (in decimal form),
   - $$t = 15$$ is the number of years.

2. Substitute the given values into the formula:
   $$
   1710 = 890 (1 + r)^{15}
   $$

3. Isolate $$(1 + r)^{15}$$ by dividing both sides by 890:
   $$
   (1 + r)^{15} = \frac{1710}{890} \approx 1.9191
   $$

4. Solve for $$(1 + r)$$ by taking the 15th root of both sides:
   $$
   1 + r = \left(1.9191\right)^{\frac{1}{15}}
   $$

5. Take the natural logarithm to compute $$r$$:
   $$
   \ln(1 + r) = \frac{1}{15} \ln(1.9191)
   $$
   Approximating:
   $$
   \ln(1.9191) \approx 0.652 \quad \Rightarrow \quad \ln(1 + r) \approx \frac{0.652}{15} \approx 0.04347
   $$
   Then,
   $$
   1 + r \approx e^{0.04347} \approx 1.0444
   $$
   So,
   $$
   r \approx 1.0444 - 1 = 0.0444
   $$

6. Convert the decimal to a percentage:
   $$
   0.0444 \times 100\% \approx 4.44\%
   $$
   Rounded to the nearest tenth of a percent, the interest rate is approximately $$4.4\%$$.
An interest rate of approximately 4.4% is needed for Joseph to grow his \$890 investment to \$1,710 over 15 years with annual compounding.

Joseph is going to invest and leave it in an account for 15 years. Assuming the
interest is compounded annually, what interest rate, to the nearest tenth of a percent,
would be required in order for Joseph to end up with ?

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Joseph is going to invest and leave it in an account for 15 years. Assuming the interest is compounded annually, what interest rate, to the nearest tenth of a percent, would be required in order for Joseph to end up with ?