A new popular midsize car has a retail price of $20,000. The midsize car's value M(t) is given by the exponential model M(t) = 20,000(4/5)^t where t represents the time in years. Identify the domain, in yearly intervals, that contains all the years the car's value is less than $1,000.

[13, ∞)
[14, ∞)
[15, ∞)
[16, ∞)

Respuesta :

Answer:

[14, ∞)

Step-by-step explanation:

M(t) = 20,000(4/5)^t

Let m(t) > 1000

1000 > 20,000(4/5)^t

Divide each side by 20000

1000/20000 > (4/5)^t

1/20 > (4/5 )^t

Take the log of each side

log (.05) > log(4/5 )^t

Using log a^b = blog a

log (.05) > t log(4/5 )

Divide each side by log (4/5)

log (.05)/ log (4/5) > t

13.42513488>t

Rounding up to the next integer

t = 14

Domain, in yearly intervals, that contains all the years the car's value is less than $1,000 for the given exponential model is [ 14, ∞).

What is exponential model?

" Exponential model is defined as the curve which represents the growth or degradation of the model as per the given condition."

According to the question,

Given exponential model,

 [tex]M(t) =20,000(\frac{4}{5} )^t[/tex]

M(t)  = Midsize car's value

t  = Time in years

Condition given to identify the domain for the given exponential model,

[tex]M(t) < 1000[/tex]

⇒[tex]20,000(\frac{4}{5} )^t < 1000[/tex]

⇒[tex](\frac{4}{5} )^t < \frac{1000}{20,000}[/tex][tex](\frac{4}{5} )^t < \frac{1}{20}[/tex]

Taking log both the sides we get,

⇒[tex]t ( log4-log5) < log1- log20[/tex]

⇒ [tex]t ( 0.6020 - 0.6989) = 1.3010[/tex]

⇒[tex]t(-0.0969 ) < (-1.3010)[/tex]

⇒[tex]t < \frac{1.3010}{0.0969}[/tex]

⇒[tex]t < 13.426[/tex]

After 13 years that is from 14 years to ∞ car's value is less than $1,000.

Therefore,

domain=[14,∞)

Hence, domain, in yearly intervals, that contains all the years the car's value is less than $1,000 for the given exponential model is [ 14, ∞).

Learn more about exponential model here

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