A person invested \$5,600$5,600 in an account growing at a rate allowing the money to double every 6 years. How long, to the nearest tenth of a year would it take for the value of the account to reach \$72,500$72,500?

22.2 years

Step-by-step explanation:

Doubling formula is given as:

P(t) = Po × (2)^t/k

Where

Po = Initial amount invested = $5600

P(t) = Amount after time t = $72500

k = Time it takes to double = 6 years

t = Time in years = ??

Hence:

72500 = 5600 × (2)^t/6

Divide both sides by 5600

72500/5600 =( 5600 × (2)^t/6)5600

12.946428571 = (2)^t/6

Take the In of both sides

In 12.946428571 = In (2)^t/6

In 12.946428571 = t/6 In (2)

Divide both sides by In 2

In 12.946428571/In 2 = t/6

3.6944822628 = t/6

Cross Multiply

t = 6 × 3.6944822628

t = 22.166893577 years

Approximately

t = 22.2 years

Therefore, it would take 22.2 years

answer: 3.26019516

step-by-step explanation:

answer: the answer is 110