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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 if 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 8, the irrational value of e³ can only be approximated in the calculator. Make sure you know how to operate your calculator for finding this type of solution. Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

• Solve log₂(x) = 4.5.

First I'll convert the log equation to the corresponding exponential form, using The Relationship:

log₂(x) = 4.5
2^4.5 = x

This requires a calculator for finding the approximate decimal value. 

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

Solving this sort of equation usually works this way: You use The Relationship to convert the log equation into the corresponding 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, don't round as you go along. Instead, do all the solving and simplification algebraically; then, at the end, do the decimal approximation as one (possibly long) set of commands in the calculator. Round-off error can get really big really fast with logs, and you don't want to lose points because you rounded too early and thus too much.

• Solve log₂(3x) = 4.5.

I'll 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 round-off-error difficulty in mind when checking your solutions; when you plug the decimal into the original equation, just make sure that the result is close enough to be reasonable. For instance, to check the solution of the equation log₂(3x) = 4.5, I'll plug 7.54 in for x:

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 darned close. Allowing for round-off error, these values confirm to me that I'd gotten the right answer.

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