Natural logarithms
Example (extension)
The power supply of a space satellite is by means of a radioisotope. The power output, in watts, is given by \({w_t} = {w_o}{e^{kt}}\) where \(t\) is the time in days. The power output at launch is 60 watts.
(a) After 14 days the power output has fallen to 56 watts. Calculate the value of \(k\) to three decimal places.
(b) The satellite cannot function properly if the power output falls below 5 watts. How many days will the satellite function properly?
(a) \({w_t} = {w_o}{e^{kt}}\)
Substitute the given values into the formula.
\(56 = 60{e^{14k}}\)
\({e^{14k}} = \frac{{56}}{{60}} = \frac{{14}}{{15}}\)
Simplify by taking the natural log of both sides.
\({\log _e}{e^{14k}} = {\log _e}\left( {\frac{{14}}{{15}}} \right)\)
\(14k{\log _e}e = {\log _e}\left( {\frac{{14}}{{15}}} \right)\)
Remember \({\log_e}e = 1\), therefore:
\(k = \frac{{{{\log }_e}\left( {\frac{{14}}{{15}}} \right)}}{{14}}\)
Use your calculator to solve:
\(k = - 0.005\,(to\,3\,d.p.)\)
Therefore the formula is now: \({w_t} = 60{e^{ - 0.005t}}\)
(b) When \({w_t} = 5\) then \(60{e^{ - 0.005t}} = 5\)
\({e^{ - 0.005t}} = \frac{5}{{60}} = \frac{1}{{12}}\)
Take the log of both sides.
\(- 0.005t{\log _e}e = {\log _e}\left( {\frac{1}{{12}}} \right)\)
\(- 0.005t = {\log _e}\left( {\frac{1}{{12}}} \right)\)
\(t = \frac{{{{\log }_e}\left( {\frac{1}{{12}}} \right)}}{{ - 0.005}}\)
Use your calculator to solve:
\(= 496.98\)
The total number of days it will take for the power output to reach 5 watts will be 496.
(Note, on the 497th day, the power output will have fallen below 5 watts.)
