Question

In 2009, the average compensation for CEOs in the U.S. was approximately $10,800,000, and by 2016, this had risen to about $12,800,000. By comparison, the average compensation for workers was $54,700 in 2009 and $55,800 in 2016. Assume that both values are growing exponentially. Find the growth rate r for both salaries. State your answers as percentages rounded to the nearest tenth of a percent.For CEOs: r= %For workers: r = %

Answer 1

The exponential growth formula is given by

`C=a(1+r)^t`

where C is the average compensation, a is the initial amount and t is the time (in years).

For the CEO's case, we have that

`\begin{gathered} \text{for t=2009, C=10800000} \\ \text{for t=2016, C=12800000} \end{gathered}`

Then, we can find a and r with these values. By substituting the first values, we have

`10800000=a(1+r)^(2009)`

By substituting the second values, we get

`128000000=a(1+r)^(2016)`

By isolating a in both case, we have that

`\begin{gathered} a=(10800000)/((1+r)^(2009)) \\ \text{and} \\ a=(12800000)/((1+r)^(2016)) \end{gathered}`

Then, we can state the following equation:

`(10800000)/((1+r)^(2009))=(12800000)/((1+r)^(2016))`

or equivalently,

`\begin{gathered} ((1+r)^(2016))/((1+r)^(2009))=(128000000)/(108000000) \\ ((1+r)^(2016))/((1+r)^(2009))=1.185185 \end{gathered}`

By applying natural logarithm in bot sides, we have

`\begin{gathered} \ln (((1+r)^(2016))/((1+r)^(2009)))=\ln 1.185185 \\ \ln (((1+r)^(2016))/((1+r)^(2009)))=0.169899 \end{gathered}`

By the properties of logarithms:

`\ln ((A)/(B))=\ln A-\ln B`

we obtain

`\ln (1+r)^(2016)-\ln (1+r)^(2009)=0.169899\ldots.(a)`

By the property of logaritm:

`\ln A^x=x\ln A`

The last result is equivalent to

`\begin{gathered} 2016\ln (1+r)-2009\ln (1+r)=0.169899 \\ which\text{ gives} \\ 7\ln (1+r)=0.169899 \end{gathered}`

then, by moving the number 7 to the right hand side, this yields,

`\begin{gathered} \ln (1+r)=(0.169899)/(7)\ldots(b) \\ \ln (1+r)=0.024271 \end{gathered}`

Now, by applying the exponential function in both sides, we have

`\begin{gathered} 1+r=e^(0.024271) \\ 1+r=1.024568 \end{gathered}`

then, r is given by

`\begin{gathered} r=1.024568-1 \\ r=0.0245 \end{gathered}`

Now, by converting this result into percent form. The answer for the CEO's case is

`r=2.45\text{ \%}`

Now, for the workers case, we can do the same procedure and get an equation similar to equation (a). That is,

`\begin{gathered} \ln (1+r)^(2016)-\ln (1+r)^(2009)=\ln ((55800)/(54700)) \\ \ln (1+r)^(2016)-\ln (1+r)^(2009)=\ln (1.020109) \\ \ln (1+r)^(2016)-\ln (1+r)^(2009)=0.0199 \end{gathered}`

and find somthing similar to equation (b)

`\begin{gathered} \ln (1+r)=(0.0199)/(7)\ldots \\ \ln (1+r)=0.0028 \end{gathered}`

and finally,

`\begin{gathered} 1+r=e^(0.0028) \\ 1+r=1.0028 \end{gathered}`

then, r will be

`\begin{gathered} r=1.0028-1 \\ r=0.0028 \\ In\text{ percent form is} \\ r=\text{ 0.28 \%} \end{gathered}`

In summary,by rounding the answers to the nearest tenth percent, the answers are

`\begin{gathered} \text{For CEO's:} \\ r=2.5\text{ \%} \\ \text{For workers:} \\ r=0.3\text{ \%} \end{gathered}`