Given,

Personnel cost \(\displaystyle=\${9},{500.000}\) in 2009

Rate of interest \(= 4.2\%\) (annually)

The annual cost function,\(\displaystyle{C}{\left({t}\right)}={9.5}{\left({1.042}\right)}^{t}\)

a)Write the equation that can be solved to find in what year personnel cost will be double the 2009 personnel costs?

b)Solve the equation numerically and determine the year.

Personnel cost will be double the 2009 personnel cost \(\displaystyle={2}\times\${9},{500},{000}= \displaystyle\${19},{000},{000}\)

So we get,

\(\displaystyle{A}={P}{\left({1}+{r}\right)}^{t}\)

\(\displaystyle{A}={9500000}{\left({1}+{0.042}\right)}^{t}\)

\(\displaystyle{19000000}={9500000}{\left({1.042}\right)}^{t}\)

\(\displaystyle{19}={9.5}{\left({1.042}\right)}^{t}\)

Find the value for t,

\(\displaystyle{199.5}={\left({1.042}\right)}^{t}\)

\(\displaystyle{2}={\left({1.042}\right)}^{t}\)

Apply logarithmic on both side we get,

\(\displaystyle \log{{2}}={ \log{{\left({1.042}\right)}}^{t}}\)

We know that, \(\displaystyle{ \log{{m}}^{n}=}{n} \log{{m}}\)

\(\displaystyle \log{{2}}={t} \log{{\left({1.042}\right)}}\)

\(\displaystyle{t}=\frac{{ \log{{2}}}}{{ \log{{\left({1.042}\right)}}}}\)

\(\displaystyle{t}=\frac{{{0.3010299}}}{{{0.0178677}}}\)

\(t = 16.8477\)

\(\displaystyle{t}\stackrel{\sim}{=}{17}\)

After 17 years we get the value of personnel cost is double in 2009.

So, \(\displaystyle{2009}+{17}={2026}\)

Therefore we get, \(\displaystyle{19}={9.5}{\left({1.042}\right)}^{t},\) Year is 2026.

Personnel cost \(\displaystyle=\${9},{500.000}\) in 2009

Rate of interest \(= 4.2\%\) (annually)

The annual cost function,\(\displaystyle{C}{\left({t}\right)}={9.5}{\left({1.042}\right)}^{t}\)

a)Write the equation that can be solved to find in what year personnel cost will be double the 2009 personnel costs?

b)Solve the equation numerically and determine the year.

Personnel cost will be double the 2009 personnel cost \(\displaystyle={2}\times\${9},{500},{000}= \displaystyle\${19},{000},{000}\)

So we get,

\(\displaystyle{A}={P}{\left({1}+{r}\right)}^{t}\)

\(\displaystyle{A}={9500000}{\left({1}+{0.042}\right)}^{t}\)

\(\displaystyle{19000000}={9500000}{\left({1.042}\right)}^{t}\)

\(\displaystyle{19}={9.5}{\left({1.042}\right)}^{t}\)

Find the value for t,

\(\displaystyle{199.5}={\left({1.042}\right)}^{t}\)

\(\displaystyle{2}={\left({1.042}\right)}^{t}\)

Apply logarithmic on both side we get,

\(\displaystyle \log{{2}}={ \log{{\left({1.042}\right)}}^{t}}\)

We know that, \(\displaystyle{ \log{{m}}^{n}=}{n} \log{{m}}\)

\(\displaystyle \log{{2}}={t} \log{{\left({1.042}\right)}}\)

\(\displaystyle{t}=\frac{{ \log{{2}}}}{{ \log{{\left({1.042}\right)}}}}\)

\(\displaystyle{t}=\frac{{{0.3010299}}}{{{0.0178677}}}\)

\(t = 16.8477\)

\(\displaystyle{t}\stackrel{\sim}{=}{17}\)

After 17 years we get the value of personnel cost is double in 2009.

So, \(\displaystyle{2009}+{17}={2026}\)

Therefore we get, \(\displaystyle{19}={9.5}{\left({1.042}\right)}^{t},\) Year is 2026.