Return to the Purplemath home page

 The Purplemath Forums
Helping students gain understanding
and self-confidence in algebra

powered by FreeFind


Return to the Lessons Index  | Do the Lessons in Order  |  Get "Purplemath on CD" for offline use  |  Print-friendly page

Solving Exponential Equations:
    Calculators and Other Considerations
(page 3 of 3)

Sections: Solving from the definition, Solving using logarithms, Calculators

You should not reduce expressions to decimal values until the very end of your computations, and then only if the decimal approximation is needed. For instance, in the last exercise on the previous page, you should not evaluate "ln350/3 )" until the very end of the exercise. You should do as much of your work as possible symbolically and "exactly"; this will go a long way toward avoiding round-off error, which (warning!) can get vary large when dealing with logarithms.

Also, be sure to "carry" as much as possible within the calculator's memory. Don't find all the values of the individual logs, write them down, re-enter them into your calculator, and then simplify; the round-off error will very likely be too large for your answer to be counted "correct". Instead, get in the habit of doing as many steps as you can all at once within the calculator's memory.

If you use a graphing calculator, typeset the entire expression at once, being careful with your parentheses. If you typeset without parentheses, as shown at right:


sloppy typesetting and wrong value

...then you will get the wrong value, because the calculator thought you meant "divide the natural log of 350 by 3, and then divide this by the log of 2, and then subtract 4"


What you need to enter is this:


correct typesetting and value

...which is "the natural log of the quotient of 350 and 3, divided by the natural log of 2, and then subtracting 4 from this". Parentheses can make a big difference on graphing calculators!

  • Solve 2ex + 5 = 115.

    I need to isolate the variable, so first I have to subtract the 5 and divide the 2 to the other side. Then I can solve by taking logs:




      2ex + 5 = 115
      x = 110
      ex = 55
      x) = ln(55)
      xln(e) = ln(55)

      x(1) = ln(55)

      x = ln(55)

    ...or about 4.007, rounded to three decimal places.

  • Solve 10000.12x = 25,000.

    Don't try to divide both sides by 1000; the 1000 is the base, not a multiplier.

    Since the base in this case is 1000, which is a power of 10, I will use the common log to solve. The natural log would have given the same answer (eventually, after some manipulation), but the base-10 log will be simpler in this case:

      10000.12x = 25,000
      log(10000.12x) = log(25,000)

      0.12xlog(1000) = log(25,000)

      0.12xlog(103) = log(25,000)

      0.12x(3) = log(25,000)

      0.36x = log(25,000)

      x = log(25,000)/0.36

    ...or about 12.217, rounded to three decimal places.

Note that the expression could have been simplified differently, by using log properties and rules) before entering it into the calculator: Copyright Elizabeth Stapel 2002-2011 All Rights Reserved

    0.36x = log(25,000)
    0.36x = log(25 1,000)

    0.36x = log(25) + log(1,000)

    0.36x = log(25) + log(103)

    0.36x = log(25) + 3

    x = [ log(25) + 3 ] / 0.36

This is equivalent to the previous form of the answer. I am pointing this out because you may need to be flexible with the form of your final answer. For instance, if you had come up with the first form of the answer, but the back of the book gave the second form of the answer, then you would need to be able to recognize that the two forms are actually the same thing. The same goes for multiple-choice tests, where your form of the answer might be different in form from, but equivalent in value to, one of the given choices.

  • Solve 250(1.04)x = 1000.

    You might recognize this as being the equation that stands for an initial investment of $250 at four percent interest, compounded annually, and asking how many years x the money should be invested in order to have $1000 in the account. To solve, I'll have to get the x by itself, so I'll divide off the 250, and then use logs:

      250(1.04)x = 1000
      x = 4
      x) = ln(4)
      xln(1.04) = ln(4)

      x = ln(4)/ln(1.04)

    ...or about 35.346, rounded to three decimal places.

If this question had been stated in terms of interest rates and investments, the above answer would have stood for "35.346 years", or about thirty-five years and four months.

As long as you do your steps clearly and completely, and keep your log rules in mind, along with the definitions of exponentials and logs, you shouldn't have too much trouble in solving these equations. Just remember to keep your work as "exact" as you can for as long as you can; wait to approximate things in your calculator until the very end, if at all possible.

<< Previous  Top  |  1 | 2 | 3  |  Return to Index

Cite this article as:

Stapel, Elizabeth. "Solving Exponential Equations: Calculators and Other Considerations."
    Purplemath. Available from


  Linking to this site
  Printing pages
  School licensing

Reviews of
Internet Sites:
   Free Help
   Et Cetera

The "Homework

Study Skills Survey

Tutoring from Purplemath
Find a local math tutor

This lesson may be printed out for your personal use.

Content copyright protected by Copyscape website plagiarism search

  Copyright 2002-2012  Elizabeth Stapel   |   About   |   Terms of Use


 Feedback   |   Error?