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 = %

Respuesta :

The exponential growth formula is given by

[tex]C=a(1+r)^t[/tex]

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

[tex]\begin{gathered} \text{for t=2009, C=10800000} \\ \text{for t=2016, C=12800000} \end{gathered}[/tex]

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

[tex]10800000=a(1+r)^{2009}[/tex]

By substituting the second values, we get

[tex]128000000=a(1+r)^{2016}[/tex]

By isolating a in both case, we have that

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

Then, we can state the following equation:

[tex]\frac{10800000}{(1+r)^{2009}}=\frac{12800000}{(1+r)^{2016}}[/tex]

or equivalently,

[tex]\begin{gathered} \frac{(1+r)^{2016}}{(1+r)^{2009}}=\frac{128000000}{108000000} \\ \frac{(1+r)^{2016}}{(1+r)^{2009}}=1.185185 \end{gathered}[/tex]

By applying natural logarithm in bot sides, we have

[tex]\begin{gathered} \ln (\frac{(1+r)^{2016}}{(1+r)^{2009}})=\ln 1.185185 \\ \ln (\frac{(1+r)^{2016}}{(1+r)^{2009}})=0.169899 \end{gathered}[/tex]

By the properties of logarithms:

[tex]\ln (\frac{A}{B})=\ln A-\ln B[/tex]

we obtain

[tex]\ln (1+r)^{2016}-\ln (1+r)^{2009}=0.169899\ldots.(a)[/tex]

By the property of logaritm:

[tex]\ln A^x=x\ln A[/tex]

The last result is equivalent to

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

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

[tex]\begin{gathered} \ln (1+r)=\frac{0.169899}{7}\ldots(b) \\ \ln (1+r)=0.024271 \end{gathered}[/tex]

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

[tex]\begin{gathered} 1+r=e^{0.024271} \\ 1+r=1.024568 \end{gathered}[/tex]

then, r is given by

[tex]\begin{gathered} r=1.024568-1 \\ r=0.0245 \end{gathered}[/tex]

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

[tex]r=2.45\text{ \%}[/tex]

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

[tex]\begin{gathered} \ln (1+r)^{2016}-\ln (1+r)^{2009}=\ln (\frac{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}[/tex]

and find somthing similar to equation (b)

[tex]\begin{gathered} \ln (1+r)=\frac{0.0199}{7}\ldots \\ \ln (1+r)=0.0028 \end{gathered}[/tex]

and finally,

[tex]\begin{gathered} 1+r=e^{0.0028} \\ 1+r=1.0028 \end{gathered}[/tex]

then, r will be

[tex]\begin{gathered} r=1.0028-1 \\ r=0.0028 \\ In\text{ percent form is} \\ r=\text{ 0.28 \%} \end{gathered}[/tex]

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

[tex]\begin{gathered} \text{For CEO's:} \\ r=2.5\text{ \%} \\ \text{For workers:} \\ r=0.3\text{ \%} \end{gathered}[/tex]