# If a boat costs \$40,000 find the number of years until the boat is worth \$18,000.  [Round the answer to the nearest tenth of a year.]A new boat will decrease in value at a rate of 8% per year...

If a boat costs \$40,000 find the number of years until the boat is worth \$18,000.  [Round the answer to the nearest tenth of a year.]

A new boat will decrease in value at a rate of 8% per year according to this formula V=C(1-r)^t  where V is the value of the boat after t years, C is the original cost, and r is the rate of depreciation. If a boat costs \$40,000 find the number of years until the boat is worth \$18,000.  [Round the answer to the nearest tenth of a year.]

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Given the formula:

V = C(1-r)^t

v is the values

c is the original cost = 40,000

r is the rate = 8% = 0.08

t is the number of years

Given the information above, we need to find t such that v= 18,000

We will substitute the given information into the equation.

==> 18,000 = 40,000 *( 1- 0.08)^t

==> 18,000 = 40,000*(0.92)^2

We will divide by 40,000

==> (0,92)^t = 0.45

Now we will take the logarithm for both sides.

==> log (0.92)^t = log 0.45

==> t* log (0.92) = log 0.45

==> t= log 0.45 / log 0.92 = 9.6 years

Then, it takes 9.6 years for the boat to be worth \$18,000.