Question

Please help with 11c!?

Answer 1

Answer:

"To the nearest year, it would be about 9 years"

Step-by-step explanation:

11c)

This is compound growth problem. It goes by the formula:

$F=P(1+r)^t$

Where

F is the future amount

P is the present (initial) amount

r is the rate of growth, in decimal

t is the time in years

Given,

P = 20,000

r = 8% = 8/100 = 0.08

F = double of initial amount = 2 * 20,000 = 40,000

We need to find t:

$F=P(1+r)^t\\40,000=20,000(1+0.08)^t\\2=(1.08)^t$

To solve exponentials, we can take Natural Log (Ln) of both sides:

$2=(1.08)^t\\Ln(2)=Ln((1.08)^t)$

Using the rule shown below we can simplify and solve:

$Ln(a^b)=bLn(a)$

We can write:

$Ln(2)=Ln((1.08)^t)\\Ln(2)=tLn(1.08)\\t=\frac{Ln(2)}{Ln(1.08)}\\t=9.0064$

To the nearest year, that would be about 9 years