Applications of logarithms

The examples below illustrate the main applications of logarithms (and exponential functions) which appear in the Higher Mathematics examination.

Given any equation of the form $y = y_0 e^{kt}$ you will either be asked to work out one of the unknowns, or to carry out a calculation involving this equation.

Have a look at the following worked example.

The power supply of a space satellite is by means of a radioisotope. The power output, in watts, is given by $w_t = w_0 e^{kt}$ where is the time in days. The power output at launch is 60 watts.

After 14 days the power output has fallen to 56 watts. Calculate the value of k to three decimal places.
The satellite cannot function properly if the power output falls below 5 watts. How many days will the satellite function properly?

First, calculate the value of to three decimal places.
$w_t = w_0 e^{kt}$
substitute the values you have into the formula
$56 = 60 e^{14k}$
${e^{14k}} = {{56} \over {60}} = {{14} \over {15}}$
simplify
$log _e e^{14k} = \log _e ({{14} \over {15}})$
$14\;k\;\log _e e = \log _e ({{14} \over {15}})$
use the laws of logarithms to separate
$k = \frac{log_e({{14} \over {15}})}{14}$
solve
k = -0.005
$w_t = w_0 e^{ - 0.005t}$ (or $w_t = 60\;e^{ - 0.005t}$ )
Secondly, work out how many days it will take for the power output to reach 5 watts.
When $w_t = 5,\;\;\;\;60\;e^{ - 0.005t} = 5$
$e^{0.005t} = {5 \over 60} = \frac{1}{12}$
${-0.005t} {log_e} {e }= log_e({1 \over 12})$
$-0.005t = log_e({1 \over 12})$
${t} = {{ log_e({1 \over 12}) } \over -0.005 }$
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.)