Question

a person invested $320 in an account growing at a rate allowing the money to double every 9 years. how long, to the nearest tenth of a year would it take for the value of the account to reach $1,010?

Explanation:

Step1: Define growth formula

For doubling-time growth, the formula is $A = P \times 2^{\frac{t}{d}}$, where $A$ = final amount, $P$ = principal, $t$ = time, $d$ = doubling time.

Step2: Plug in given values

Substitute $A=1010$, $P=320$, $d=9$:
$1010 = 320 \times 2^{\frac{t}{9}}$

Step3: Isolate the exponential term

Divide both sides by 320:
$\frac{1010}{320} = 2^{\frac{t}{9}}$
Simplify: $3.15625 = 2^{\frac{t}{9}}$

Step4: Take log of both sides

Use natural log: $\ln(3.15625) = \frac{t}{9} \times \ln(2)$

Step5: Solve for t

Rearrange to solve for $t$:
$t = 9 \times \frac{\ln(3.15625)}{\ln(2)}$
Calculate: $\ln(3.15625) \approx 1.148$, $\ln(2) \approx 0.693$
$t \approx 9 \times \frac{1.148}{0.693} \approx 9 \times 1.657 \approx 14.9$

Answer:

14.9 years