Print-friendly page

Solving Logarithmic Equations:
     Calculator Considerations (page 3 of 3)

Sections: Solving from the definition, Solving with exponentials, Calculator Considerations

The next level of this type of log equation may require a calculator to solve. An example would be:

• Solve ln(x) = 3.

The base of the natural logarithm is the number "e" (with a value of about 2.7).  To solve this, I'll use The Relationship, and I'll keep in mind that the base is "e":

ln(x) = 3
e³ = x

This is a valid solution, and may be all that your books wants for the answer. However, if you need to graph this value or this is the answer to a word problem, then a decimal approximation may be more useful. In that case, the answer is x = 20.086, rounded to three decimal places.

Note that this decimal form is not "better" than "e³"; actually, "e³" is the exact, and therefore the more correct, answer. But whereas something like "2³" can be simplified to a straight-forward number ("8"), "e³" must be done in the calculator. Make sure you know how to operate your calculator for finding this type of solution. Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

• Solve log₂(x) = 4.5.

Convert, using The Relationship:

log₂(x) = 4.5
2^4.5 = x

This requires a calculator for simplification. 

The answer is x = 2^4.5 = 22.63, rounded to two decimal places.

These problems usually work this way: You use The Relationship to convert the log equation into an exponential equation, and then you may or may not need to use your calculator to find an approximation of the exact form of the answer.

By the way, when finding approximations with your calculator, try not to round as you go along. Instead, try to do all the simplification algebraically, and then do the approximation as one (possibly long) set of commands in the calculator.  Round-off error can get really big really quickly with logs, and you don't want to lose points because you rounded too early.

• Solve log₂(3x) = 4.5.

Use The Relationship, followed by some algebra:

log₂(3x) = 4.5
2^4.5 = 3x
(2^4.5) ÷ 3 = x
x = 7.54247233266...

The answer is x = (2^4.5) / 3, or x = 7.54, rounded to two decimal places.

If you try to check your solution by plugging "7.54" into the calculator for "x" in the original equation, you will get a result that is close to 4.5, but not exact. This will be due to round-off error. Keep this in mind, and either plug in the exact value for x, or else just check that the result is close enough to be reasonable:

log₂(3x) = 4.5
log₂(3(7.54)) ?=? 4.5
log₂(22.62) ?=? 4.5
ln(22.62) ÷ ln(2) ?=? 4.5
4.49952702422... ?=? 4.5

Okay, they're not equal, but they're pretty close, so, allowing for round-off error, this check makes me believe that I got the right answer.

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

  Copyright © 2006-2008  Elizabeth Stapel   |   About   |   Terms of Use

 Feedback   |   Error?