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:
    Solving from the& Definition
(page 1 of 3)

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

To solve exponential equations without logarithms, you need to have equations with comparable exponential expressions on either side of the "equals" sign, so you can compare the powers and solve. In other words, you have to have "(some base) to (some power) equals (the same base) to (some other power)", where you set the two powers equal to each other, and solve the resulting equation. For example:

  • Solve 5x = 53.

    Since the bases ("5" in each case) are the same, then the only way the two expressions could be equal is for the powers also to be the same. That is:

      x = 3

This solution demonstrates how this entire class of equation is solved: if the bases are the same, then the powers must also be the same, in order for the two sides of the equation to be equal to each other. Since the powers must be the same, then you can set the two powers equal to each other, and solve the resulting equation.




  • Solve 101x = 104

    Since the bases are the same, I can equate the powers and solve:

      1 x = 4
      1 4 = x

      3 = x

Sometimes you'll first need to convert one side or the other (or both) to some other base before you can set the powers equal to each other. For example:

  • Solve 3x = 9. Copyright Elizabeth Stapel 2002-2011 All Rights Reserved

    Since 9 = 32, this is really asking me to solve:

      3x = 32

    By converting the 9 to a 32, I've converted the right-hand side of the equation to having the same base as the left-hand side. Since the bases are now the same, I can set the two powers equal to each other:

      x = 2

  • Solve 32x1 = 27.

    In this case, I have an exponential on one side of the "equals" and a number on the other. I can solve the equation if I can express the "27" as a power of 3. Since 27 = 33, then I can convert and proceed with the solution:

      32x1 = 27
      2x1 = 33
      2x 1 = 3

      2x = 4

      x = 2

As you can probably tell, you will need to get good with your powers of numbers, such as the powers of 2 up through 26 = 64, the powers of 3 up through 35 = 243, the powers of 4 up through 44 = 256, the powers of 5 up through 54 = 625, the powers of 6 up through 63 = 216, and all the squares. Warning: Don't plan to depend on your calculator for everything, because having to find every value in your calculator can waste a lot of time. You'll want to have a certain degree of facility, of familiarity and speed, by the time you reach the test, so familiarize yourself with the smaller powers now.

  • Solve 3x^23x = 81.

    Formatting note: HTML doesn't generally "like" nested superscripts, so the above uses the "carat" notation to denote the exponent.

    This exercise works just like the previous one:

      3x^23x = 81
      x^23x = 34
      x2 3x = 4

      x2 3x 4 = 0

      (x 4)(x + 1) = 0

      x = 1, 4

  • Solve 42x^2+2x = 8.

    This equation is similar to the previous two but is not quite the same, because 8 is not a power of 4. However, both 8 and 4 are powers of 2, so I can convert:

      4 = 22
      8 = 23

      2x^2+2x = (22)2x^2+2x = 2(2)(2x^2+2x) = 24x^2+4x

    Now I can solve:

      42x^2+2x = 8
      4x^2+4x = 23
      4x2 + 4x = 3

      4x2 + 4x 3 = 0

      (2x 1)(2x + 3) = 0

      x = 1/2 , 3/2

  • Solve 4x+1 = 1/64.

    Negative exponents can be used to indicate that the base belongs on the other side of the fraction line. Since 64 = 43, then I can use negative exponents to convert the fraction to an exponential expression: 1/64 = (43)1 = 43. Using this, I can solve the equation:

      4x+1 = 1/64
      x+1 = 43
      x + 1 = 3

      x = 4

  • Solve 8 x2 = sqrt[8]

    I need to recall that square roots are the same as one-half powers, and convert the radical to exponential form. Then I can solve the equation:

      8 x2  =  sqrt[8]
      x2  =  8 1/2
      x  2  =  1/2
      x =  2 1/2  =  5/2

Warning: The following is an example of a common type of trick question:

  • Solve 2x = 4
  • Think about it: What power on the positive number "2" could possibly yield a negative number? A number can never go from positive to negative by taking powers; I can never turn a positive two into a negative anything, four or otherwise, by multiplying two by itself, regardless of the number of times I do the multiplication. Exponentiation simply doesn't work that way. So the answer here is:

      no solution

Top  |  1 | 2 | 3  |  Return to Index  Next >>

Cite this article as:

Stapel, Elizabeth. "Solving Exponential Equations From the Definition." Purplemath. Available from Accessed


  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?