Chapter 7.1, Problem 60PPE

To determine

To find: The club’s budget $2\text{yr}$ ago, the expression models for the budget of the club in $1\text{yr}$ , $2\text{yr}$ and in $y\text{yr}$ ago and the value of $y$ to find the budget of the club $2\text{yr}$ ago.

Answer to Problem 60PPE

The club’s budget $2\text{yr}$ ago is $\$125$ , the expression for budget of the club in $1\text{yr}$ , $2\text{yr}$ and $y\text{yr}$ are $\$500\times 2$ , $\$500\times 2^2$ and $\$500\times 2^y$ respectively and the value of $y\text{yr}$ for $2\text{yr}$ ago is $-2$ .

Explanation of Solution

Given information:

The club’s budget doubles every year and the budget of this year is $\$500$ .

Calculation:

The given budget of this year is:

  $\$500$

As club budget doubles every year, so $2\text{yr}$ ago the current budget reduces by 4 times. Therefore, the drama club’s budget $2\text{yr}$ ago is:

  $\frac{\$500}{4}$

The expression for club’s budget when it doubles for $n$ year is:

  $\$500\times 2^n$

To determine the budget of the club for $1$ year, substitute $1$ for $n$ in the above expression.

  $\$500\times 2^1=\$500\times 2$

To determine the budget of the club for $2$ year, substitute $2$ for $n$ in the above expression.

  $\begin{array}{c}\$500\times 2^2=\$500\times 2\times 2\\ =\$500\times 4\end{array}$

To determine the budget of the club for $y$ year, substitute $y$ for $n$ in the above expression.

  $\$500\times 2^y$

The budget of the club for $2$ year ago from the above expression is:

  $\begin{array}{c}\$500\times 2^y=\$125\\ 2^y=\frac{\$125}{\$500}\end{array}$

Divide the numerator and denominator at the right hand side of the above expression by $125$ .

  $\begin{array}{c}{2}^y=\frac{\$125÷125}{\$500÷125}\\ =\frac{1}{4}\\ =\frac{1}{2^2}\\ =2^{-2}\end{array}$

On comparing both the sides of the above expression:

  $y=-2$

Therefore, the club’s budget $2\text{yr}$ ago is $\$125$ , the expression for budget of the club in $1\text{yr}$ , $2\text{yr}$ and $y\text{yr}$ are $\$500\times 2$ , $\$500\times 2^2$ and $\$500\times 2^y$ respectively and the value of $y$ for $2\text{yr}$ ago is $-2$ .