Question

Jack bought a new boat for £12 500.
The value, £V, of Jack's boat at the end of n years is given by the formula
V = 12 500 × (0.85)^n
(a) At the end of how many years was the value of Jack's boat first less than 50% of the value of the boat when it was new?

Answer 1

To find the number of years when the value of Jack's boat first becomes less than 50% of its original value, we can set up an equation using the given formula:

V = 12,500 × (0.85)^n

We want to find the value of n when V is less than 50% of the original value, which is 12,500 / 2 = 6,250.

So, the equation becomes:

6,250 = 12,500 × (0.85)^n

To solve for n, we need to take the logarithm of both sides of the equation. Let's use the natural logarithm (ln) for this calculation:

ln(6,250) = ln(12,500 × (0.85)^n)

Using logarithm properties, we can simplify the equation:

ln(6,250) = ln(12,500) + ln(0.85)^n

ln(6,250) = ln(12,500) + n * ln(0.85)

Now, we can solve for n by isolating it on one side of the equation:

n = (ln(6,250) - ln(12,500)) / ln(0.85)

Using a calculator, we can find the value of n.

Answer 2

Step-by-step explanation:

V(n) = 12500 × (0.85)^n

it means that it depreciates (100 - 85 = 15%) every year.

the number of years when its value reaches below 50% of the initial price (= 12500/2 = 6250) is the lowest value of n for which the following inequality is true :

6250 > 12500 × (0.85)^n

to address this problem let's deal with the line limit (equation) of the inequality area first :

6250 = 12500 × (0.85)^n

6250/12500 = (0.85)^n

0.5 = (0.85)^n

to solve this we need to use a logarithm.

e.g. the logarithm to the base of 10

log(0.5) = log((0.85)^n)

remember,

loga(b^c) = c×loga(b)

so,

log(0.5) = n × log(0.85)

n = log(0.5)/log(0.85) = -0.301029996.../-0.070581074... =

= 4.265024282... years

since we are asked for the end of how many full years, we need to round our answer up to the next full year.

which means after 5 years.