Finding x from given y value

1. The value of a car is modelled by the formula

$$V = 16000 \mathrm { e } ^ { - k t } + A , \quad t \geqslant 0 , t \in \mathbb { R }$$ where \(V\) is the value of the car in pounds, \(t\) is the age of the car in years, and \(k\) and \(A\) are positive constants. Given that the value of the car is \(\pounds 17500\) when new and \(\pounds 13500\) two years later,

1. find the value of \(A\),
2. show that \(k = \ln \left( \frac { 2 } { \sqrt { 3 } } \right)\)
3. Find the age of the car, in years, when the value of the car is \(\pounds 6000\) Give your answer to 2 decimal places.

1. The value, \(\pounds V\), of a vintage car \(t\) years after it was first valued on 1 st January 2001, is modelled by the equation

$$V = A p ^ { t } \quad \text { where } A \text { and } p \text { are constants }$$ Given that the value of the car was \(\pounds 32000\) on 1st January 2005 and \(\pounds 50000\) on 1st January 2012

1. 1. find \(p\) to 4 decimal places,
   2. show that \(A\) is approximately 24800
2. With reference to the model, interpret
   1. the value of the constant \(A\),
   2. the value of the constant \(p\). Using the model,
3. find the year during which the value of the car first exceeds \(\pounds 100000\)

1. The population in thousands, \(P\), of a town at time \(t\) years after \(1 ^ { \text {st } }\) January 1980 is modelled by the formula

$$P = 30 + 50 \mathrm { e } ^ { 0.002 t }$$ Use this model to estimate

1. the population of the town on \(1 { } ^ { \text {st } }\) January 2010,
2. the year in which the population first exceeds 84000 . The population in thousands, \(Q\), of another town is modelled by the formula $$Q = 26 + 50 \mathrm { e } ^ { 0.003 t }$$
3. Show that the value of \(t\) when \(P = Q\) is a solution of the equation $$t = 1000 \ln \left( 1 + 0.08 \mathrm { e } ^ { - 0.002 t } \right) .$$
4. Use the iteration formula $$t _ { n + 1 } = 1000 \ln \left( 1 + 0.08 \mathrm { e } ^ { - 0.002 t _ { n } } \right)$$ with \(t _ { 0 } = 50\) to find \(t _ { 1 } , t _ { 2 }\) and \(t _ { 3 }\) and hence, the year in which the populations of these two towns will be equal according to these models.

5. The number of bacteria present in a culture at time \(t\) hours is modelled by the continuous variable \(N\) and the relationship $$N = 2000 \mathrm { e } ^ { k t } ,$$ where \(k\) is a constant. Given that when \(t = 3 , N = 18000\), find

1. the value of \(k\) to 3 significant figures,
2. how long it takes for the number of bacteria present to double, giving your answer to the nearest minute,
3. the rate at which the number of bacteria is increasing when \(t = 3\).

4 A scientist is testing models for the growth and decay of colonies of bacteria. For a particular colony, which is growing, the model is \(P = A \mathrm { e } ^ { \frac { 1 } { 8 } t }\), where \(P\) is the number of bacteria after a time \(t\) minutes and \(A\) is a constant.

1. This growing colony consists initially of 500 bacteria. Calculate the number of bacteria, according to the model, after one hour. Give your answer to the nearest thousand.
2. For a second colony, which is decaying, the model is \(Q = 500000 \mathrm { e } ^ { - \frac { 1 } { 8 } t }\), where \(Q\) is the number of bacteria after a time \(t\) minutes. Initially, the growing colony has 500 bacteria and, at the same time, the decaying colony has 500000 bacteria.
   1. Find the time at which the populations of the two colonies will be equal, giving your answer to the nearest 0.1 of a minute.
   2. The population of the growing colony will exceed that of the decaying colony by 45000 bacteria at time \(T\) minutes. Show that $$\left( \mathrm { e } ^ { \frac { 1 } { 8 } T } \right) ^ { 2 } - 90 \mathrm { e } ^ { \frac { 1 } { 8 } T } - 1000 = 0$$ and hence find the value of \(T\), giving your answer to one decimal place.
      (4 marks)

4 A painting was valued on 1 April 2001 at \(\pounds 5000\).
The value of this painting is modelled by $$V = A p ^ { t }$$ where \(\pounds V\) is the value \(t\) years after 1 April 2001, and \(A\) and \(p\) are constants.

1. Write down the value of \(A\).
2. According to the model, the value of this painting on 1 April 2011 was \(\pounds 25000\). Using this model:
   1. show that \(p ^ { 10 } = 5\);
   2. use logarithms to find the year in which the painting will be valued at \(\pounds 75000\).
3. A painting by another artist was valued at \(\pounds 2500\) on 1 April 1991. The value of this painting is modelled by $$W = 2500 q ^ { t }$$ where \(\pounds W\) is the value \(t\) years after 1 April 1991, and \(q\) is a constant.
   1. Show that, according to the two models, the value of the two paintings will be the same \(T\) years after 1 April 1991, $$\text { where } T = \frac { \ln \left( \frac { 5 } { 2 } \right) } { \ln \left( \frac { p } { q } \right) }$$
   2. Given that \(p = 1.029 q\), find the year in which the two paintings will have the same value.
      [0pt] [1 mark]

6 An on-line science website states:
'To find a dog's equivalent human age in years, multiply the natural logarithm of the dog's age in years by 16 then add 31.' 6

1. Calculate the equivalent age to the nearest human year of a dog aged 5 years. 6
2. A dog's equivalent age in human years is 40 years. Find the dog's actual age to the nearest month.
   6
3. Explain why the behaviour of the natural logarithm for values close to zero means that the formula given on the website cannot be true for very young dogs.