Introduction to Logarithms: Solutions

Here are some solutions to the Exercises to accompany the section Introduction to Logarithms.  

Introduction to Logarithmic Functions

  1. Use the definition of logarithms to express the solutions to the following exponential equations:
    1. 20 = 5x
      Solution
      Since the base is 5, we must use log5.  Since we want the input of the exponential, exp5(x) = 5x, corresponding to the value 20, by the definition, the value of the desired exponent is x = log5(20).
    2. 4 = (0.75)x
      Solution
      As before, the solution to this equation is x = log0.75(4).  Notice how the base (i.e., 0.75) is the same in the equation we are trying to solve, and the logarithmic function we need to use (i.e., log0.75).  Also, the value of the logarithm is the desired exponent.
    3. 0.35 = (0.5)x
      Solution
      By the definition, the solution to this equation is x = log0.5(0.35).  Since log0.5 is the inverse of (0.5)x, it take the given output (i.e., 0.35) of (0.5)x back to the corresponding input (i.e., the desired solution, x).
    4. 0.343 = (0.49)x
      Solution
      By the definition, the solution to this equation is x = log0.49(0.343).  
    5.  -14 = -3·2x + 1.  Hint: Solve the equation first to isolate the exponential, 2x, then solve as before.
      Solution
      Subtracting -1 and dividing by -3 gives (-14 - 1)/(-3) = 2x, or 5 = 2x.  Therefore, the solution is x = log2(5).

    Back to Exercises.

  2. In the following applet, we have defined the functions f(x) = 5x, s(x) = (0.75)x, g(x) = (0.5)x, h(x) = (0.49)x, and k(x) = -3·2x + 1:

    Use the "x = " box to estimate values, to three decimals, for the solutions (expressed in terms of logarithms) in the previous Exercise.

    Solution
    1. 20 = 5x: Since 25 = 52, and the graph is increasing, we know that the solution is a little less than x = 2.  Clicking with the mouse directly on the graph gives a better estimate of x = 1.85.  Experimenting with the "x = " box gives:
      x f(x) = 5x
      1.85
      1.86
      1.87
      1.88      		 19.6379...
      19.9565...
      20.2803...
      20.6093...

      This shows that log5(20) » 1.86 is the best estimate to three digits.

    2. 4 = (0.75)x: Clicking with the mouse directly on the graph gives an estimate of x = -4.8.  Experimenting with the "x = " box gives:
      x s(x) = (0.75)x
      -4.8
      -4.81
      -4.82
      -4.83      		 3.97838...
      3.98984...
      4.00133...
      4.01286...

      This shows that log0.75(4) » -4.82 is the best estimate to three digits.

    3. 0.35 = (0.5)x: Since 0.25 = (0.5)2, and the graph is decreasing, we know that the solution is a little less than x = 2.  Clicking with the mouse directly on the graph gives a better estimate of x = 1.5.  Experimenting with the "x = " box gives:
      x g(x) = (0.5)x
      1.50
      1.51
      1.52
      1.53      		 0.353553...
      0.351111...
      0.348686...
      0.346277...

      This shows that log0.5(0.35) » 1.51 is the best estimate to three digits.

    4. 0.343 = (0.49)x: Since 0.343 » 0.35 and  0.5 » 0.49, we would expect the solution to be close to that of part c), namely x = 1.51.  Experimenting with the "x = " box shows that x = 1.5 seems to give exactly the right answer.  This is, in fact, the exact solution, which we can verify by rules of exponents:

      Thus, we can know that the solution to 0.343 = (0.49)x is precisely 1.5 = log0.49(0.343).

    5. -14 = -3·2x + 1: From the previous Exercise we know that the solution is the same as that of 5 = 2x, that is, x = log2(5).  We can easily estimate the value to be between 2 and 3.  Clicking with the mouse directly on the graph gives an estimate of x = 2.3.  Experimenting with the "x = " box gives:
      x k(x) = -3·2x + 1
      2.30
      2.31
      2.32
      2.33      		 4.92458...
      4.95883...
      4.99332...
      5.02805...

      This shows that log2(5) » 2.33 is the best estimate to three digits.

    Back to Exercises.

Graphing Logarithmic Functions

  1. Create plots for each of the following pair of exponential and logarithmic functions:
    1. exp3(x) = 3x and log3(x).
      Solution
      We can make the table for the exponential function, exp3:
      x y = exp3(x)
      -„
      -1
      0
      1      		 0
      1/3
      1
      3

      Since the base is greater than 1, we know that this goes up from the asymptote as we move to the right.  Reversing the values gives a table for the logarithm, log3:

      x y = log3(x)
      0
      1/3
      1
      3      		 -„
      -1
      0
      1

      Plotting the vertical asymptote, reversing the roles of the horizontal and vertical, we know that the graph must go up from the asymptote as we move to the right:

    2. exp1/4(x) = (1/4)x and log1/4(x).
      Solution
      As before, we first make a table for the exponential function, exp1/4:
      x y = exp1/4(x)
      -1
      0
      1
      „       		1/(1/4) = 4
      1
      1/4
      0

      This time the base is less than 1, so we know that this goes down to the asymptote as we move to the right.  Reversing the values gives a table for the logarithm, log1/4:

      x y = log1/4(x)
      4
      1
      1/4
      0      		 -1
      0
      1
      „

      Plotting the vertical asymptote, this time the graph must go down from the asymptote as we move to the right:

    3. Pick a base, b, and plot expb(x) = bx and logb(x) for your chosen base.  Repeat this Exercise as often as necessary until you are confident in your ability to plot logarithmic graphs.
      Solution
      Once you have created your own graph, you can check your work by entering your base in the following box and clicking the "Set Base" button.

      Select the base, b =

      Compare your graph with that of "logb" in the corresponding applet.  Use the "x =" box to check the points that you plotted to construct your graph.  

    Back to Exercises.

  2. Use our transformational graphing technique to graph the following functions with a logarithmic "core".  Hint: Start from the graphs from the previous Exercise:
    1. f(x) = 2log3(-x + 1) - 4.
      Solution
      From the previous Exercise, we already have a table for the "core" function, log3:
      x y = log3(x)
      0
      1/3
      1
      3      		 -„
      -1
      0
      1

      and a sketch:

      We can decompose f as:

      • Negate the input,
      • add 1,
      • apply log3,
      • multiply by 2, and
      • subtract 4.

      This leads to the following analysis:

      Corresponding
      Algebraic formula      		 log3(x) log3(x + 1) log3(-x + 1)   2log3(-x + 1) 2log3(-x + 1) - 4
      Geometric Effect Take the logarithm
      function with base 3      		 Shift left 1 Flip
      horizontally      		 Stretch vertically
      by a factor of 2      		 Shift down 
      4 units
      Numerical Effect   Subtract 1 Negate Apply the logarithm
      function with base 3      		 Multiply by 2 Subtract 4
      Numerical 
      Results      		 Inputs Outputs
      0 -1 1 -„ -„ -„
      1/3 -2/3 2/3 -1 -2 -6
      1 0 0 0 0 -4
      3 2 -2 1 2 -2

      Shifting left will move the vertical asymptote to x = -1.  Flipping the graph horizontally, will flip the asymptote over to x = 1, and the graph will now decrease as it approaches the asymptote from the left.  Stretching and shifting vertically will not have much of an effect on the graphs shape, except to make it a bit steeper.  Drawing in the asymptote, plotting the points and "connecting-the-dots" gives:

       

      You can see how this graph is "built" out of the graph of log3 by selecting the Examples in the following applet:

    2. g(x) = 3log1/4(x - 2) + 1.
      Solution
      Using the table:
      x y = log1/4(x)
      4
      1
      1/4
      0      		 -1
      0
      1
      „

      and sketch:

      from the previous Exercise for the "core" function, log1/4, we then decompose g as:

      • Subtract 2 from the input,
      • apply log1/4,
      • multiply by 3, and
      • add 1,

      which leads to the following analysis:

      Corresponding
      Algebraic formula      		 log1/4(x) log1/4(x - 2)   3log1/4(x - 2) 3log1/4(x - 2) + 1
      Geometric Effect Take the logarithm
      function with base 1/4      		 Shift right 2 Stretch vertically
      by a factor of 3      		 Shift up 
      1 unit
      Numerical Effect   Add 2 Apply the logarithm
      function with base 1/4      		 Multiply by 3 Add 1
      Numerical 
      Results      		 Inputs Outputs
      0 2 „ „ „
      1/4 9/4 1 3 4
      1 3 0 0 1
      4 6 -1 -3 -2

      Shifting right moves the vertical asymptote to x = 2.  Stretching and shifting vertically will not have much of an effect on the graphs shape, except to make it a bit steeper.  Drawing in the asymptote, plotting the points and "connecting-the-dots" gives:

       

      You can see how this graph is "built" out of the graph of log1/4 by selecting the Examples in the following applet:

    3. Fill-in-the-blanks to create a formula for a function with a logarithmic "core", then use our transformational graphing technique to graph it.

      y = __·log_(__x + __) + __

      Note: This includes choosing the base of the logarithm; you may want to use the same base that you chose in the previous Exercise.

      Repeat this Exercise as often as necessary until you are confident in your ability to plot functions with a logarithmic "core".
      Solution
      You may check your work by first entering the base that you used in the following box and clicking the "Set Base" button.

      Select the base, b =

      Then, define a new function with your formula, using the "logb" function.  Verify that its graph looks like the graph you constructed by hand.  Also, make sure to use the "x = " box to check that it goes through the points you plotted by hand.

    Back to Exercises.

Go to Properties of Logarithms.

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