Question

type the correct answer in the box. use numerals instead of words. if necessary, use / for the fraction bar.
david invested $220 in a savings account that offers a 3% return on the investment. the value of davids investment will be at least $400 after a period of

years.
hint: use the formula $a = p(1 + r)^t$, where $a$ is the amount after $t$ years, $p$ is the amount invested, $r$ is the rate of interest, and $t$ is the time period. use a calculator to compute the answer, and round it off to the nearest year.

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Explanation:

Step1: Identify given values

We know \( P = 2220 \), \( r = 0.03 \) (since 3% = 0.03), and \( A = 4000 \) (assuming the target amount is $4000, as the text seems to have a typo, likely "at least $4000"). The formula is \( A = P(1 + r)^t \).

Step2: Substitute values into formula

Substitute \( A = 4000 \), \( P = 2220 \), \( r = 0.03 \) into \( A = P(1 + r)^t \):
\( 4000 = 2220(1 + 0.03)^t \)
Simplify \( 1 + 0.03 = 1.03 \), so \( 4000 = 2220(1.03)^t \).
Divide both sides by 2220: \( \frac{4000}{2220} = (1.03)^t \)
Simplify \( \frac{4000}{2220} \approx 1.8018 \), so \( 1.8018 = (1.03)^t \).

Step3: Solve for \( t \) using logarithms

Take the natural logarithm of both sides: \( \ln(1.8018) = \ln((1.03)^t) \)
Using the logarithm power rule \( \ln(a^b) = b\ln(a) \), we get:
\( \ln(1.8018) = t\ln(1.03) \)
Solve for \( t \): \( t = \frac{\ln(1.8018)}{\ln(1.03)} \)
Calculate \( \ln(1.8018) \approx 0.588 \) and \( \ln(1.03) \approx 0.02956 \).
Then \( t \approx \frac{0.588}{0.02956} \approx 19.9 \), which rounds to 20.

Answer:

20