Solving equations using logs

This section explores solving equations using logs. The section covers solving equations and inequalities usig logs, fiding intersection s of exponential graphs uing logs. Before atempting this chapter you must have prior knowledge of logarithms basics and rearranging equations

Solve equations using logs on both sides

Let us consider the following formula for the growth of aliens.

y = 2x

…where y is the number of aliens and x is the number of days.

We can ue logs to change the subject of the formula to find x for example.

Example:Suppose we wanted to find how long it takes before there are 1000000 aliens.

We substitute in the value to form the equation shown below.

2x = 1000 000
Answer:
Explanation: Now we need to rearrange it to make x the subject.

log xx = nlogx

You may not be able to find log2 10000000 on the calculator, a calculator may have log10 and not log2. So instead we take logs on both sides to base 10 we use;

log 2x = 1000 000

So now we have;

xlog 2 = log 1000 000

…now we divide both sides by log 2 to get;

x = log 1000 000/log 2

Now it is easier to use the calculator to work it out.

When worked out we see that;

x = 6/0.3010
x – 19.9

…round it;

x = 20

We can check it in the original formula to see whether the answer is correct.

220 = 1048576

Example:Solve the following expression by finding the value of x.

42x + 1 = 60
Answer:

    Explanation:Here is another example. Suppose we wanted to solve;

    …again we need to use;

    log xn = nlogx

    …to get;

    log 42x + 1 = log 60

    …simplify…

    (2x + 1)log4 = log 60
    2x + 1 = log 60/log 4
    2x + 1 = 2.95
    2x = 1.95

    Solving inequalities

    Example:

    Answer:

      Explanation:
      We can also solve inequalities… Suppose we wanted to find the smallest integer x such that;

      0.3x < 0.001

      …we use logs to get;

      log 0.3x < log 0.001

      …we use;

      log xn = nlog x

      to get;

      xlog 0.3 < log 0.001

      We have to reverse the inequality sign because log 0.3 is negative.

      x > log 0.001/log 0.3
      x > 7.65

      We can conclude that the smallest integer must be 8.

      Finding intersections of log graphs

      We can also find intersection of log graphs.

      Example: Suppose we wanted to find the x coordinate of the point of intersection for the following curves.

      y = 2-x and y = 3x + 1
      Answer:

        Explanation:
        Also suppose we wanted to sketch the graphs showing the intersection.

        Normally to find the intersection we need to put the equations equal so we have.

        2-x = 3x + 1

        …use logs to solve the equations

        log 2-x = log 3x + 1

        …we rearrange the equation using;

        log xn = nlogx

        …to get;

        -xlog 2 = (x + 1)log 3

        …we divide through by 3

        -xlog 2/log 3 = x + 1

        …we work out the log division

        -x × 0.631 = x + 1
        -0.631 x = x + 1

        …now we simplify.

        -1 = x + 0.631x
        -x = 1.631x
        x = -1/1.631 = -0.613

        The curves intersect at x = -0.613

        Now we can sketch the graph, first we sketch

        y = 2-x

        The graph is simply y=2x reflected in the y axis.
        [graph id=”26″]
        Next we sketch 3x + 1
        [graph id=”27″]
        On the graph we can see that the curves intersect at;

        x = -0.6

        Harder equations involving logs

        Below we shall explore another equation involving logs.

        Example:Suppose we had to solve.

        1 + 2 log3x = log3 (5x + 2)
        Answer:

          Explanation:
          The problem looks complicated. First we take all the terms to one side.

          2 log3 x – log3 (5x + 2) = -1

          Now we can use;

          log xx = nlog x

          …to get;

          log3 x2 – log3 (5x + 2) = -1

          …we use the subtraction rule to get;

          log3 x/5x + 2 = -1

          By doing a power 3 on the opposite side we can remove log3

          x2/5x + 2 = 3-1 = 1/3

          …we simplify to get;

          3x2 = 5x + 2

          …then we rearrange to get;

          3x2 – 5x – 2 = 0

          Now we can factorise and solve;

          (3x + 1)(x – 2) = 0
          x = -1/3 or x = 2

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