The
Purplemath Forums 
Solving Logarithmic
Equations: 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:
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 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 "e^{3}"; actually, "e^{3}" is the exact, and therefore the more correct, answer. But whereas something like 2^{3} can be simplified to a straightforward 8, the irrational value of e^{3} 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 20022011 All Rights Reserved
First I'll convert the log equation to the corresponding exponential form, using The Relationship: log_{2}(x)
= 4.5
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. Roundoff 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.
I'll use The Relationship, followed by some algebra: log_{2}(3x)
= 4.5
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 roundoff error. Keep this roundofferror 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 log_{2}(3x) = 4.5, I'll plug 7.54 in for x: log_{2}(3x)
= 4.5 Okay, they're not equal, but they're pretty darned close. Allowing for roundoff error, these values confirm to me that I'd gotten the right answer. << Previous Top  1  2  3  Return to Index



Copyright © 20022012 Elizabeth Stapel  About  Terms of Use 




