v(t)=18 500e^(- 0.186t)
The number v(t) is the value of the car after t years. We are asked to find the worth of the car after 18 months, i.e., after 1.5 years. Therefore, we will substitute 1.5 for t into the given equation.
To find when the value of the car will be halved, first let's find the original value of the car. To do this we will substitute t=0 into the given formula.
To solve the equation for t we will use the definition of a logarithm.
x= b^y ⇔ log_b x=y
This definition tells us how to rewrite the logarithm equivalent of y as an exponential equation. The argument x is equal to b raised to the power of y. Let's do it!
1/2= e^(- 0.186 t) ⇔ log_e 1/2=- 0.186 t
Notice also that log_e x is equal to ln x for all x.
log_e1/2=- 0.186 t ⇔ ln1/2=- 0.186 t
Finally, let's solve the last equation for t.
To solve the equation for t we will use the definition of a logarithm.
x= b^y ⇔ log_b x=y
This definition tells us how to rewrite the logarithm equivalent of y as an exponential equation. The argument x is equal to b raised to the power of y. Let's do it!
2/37= e^(- 0.186 t) ⇔ log_e 2/37=- 0.186 t
Notice also that log_e x is equal to ln x for all x.
log_e2/37=- 0.186 t ⇔ ln2/37=- 0.186 t
Finally, let's solve the last equation for t.
