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Maths

Logarithms

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 t is the time in days. The power output at launch is 60 watts.

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

    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})

     

  2. 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 }

    = 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.)

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