Simple Interest: I and Compounded Interest: A

Question

Ryan bought a new computer for $3,400. The value of the computer decreases 50% each year. when will the value drop below $300?

Answer 1

It drops below $300 at the end of 4 years!

You can find that by just assessing the value at the end of each year!

End of Year 1: goes from $3400 to $1700

End of Year 2: goes from $1700 to $850

End of Year 3: goes from $850 to $425

End of Year 4: goes from $425 to $212.50

so in between year 3 and 4 is when it hits the $300 price point

Answer 2

The general form for simple compounding interest is f = i (1 + r)^t, where f is the final value, i is the initial value, r is the annual rate of growth (or negative if decay), and t is time in years. From the problem prompt, the final value after some time f = 300, the initial value i = 3400, the annual growth rate i = -0.5 (negative because "decreases", and 0.5 because 50%), and t is what we are solving for. Rearranging the equation for t and substituting our values:

f = i (1 + r)^t

(f / i) = (1 + r)^t

ln(f / i) = t ln(1 + r) where ln(x) is the natural log function of x, or log_e(x)

t = ln(f / i) / ln(1 + r)

t = ln(300 / 3400) / ln(1 - 0.5) by substitution, then put into calculator

t = 3.5 <<< final answer, in unit of years to 1 decimal place.