The next level of this type of log equation may require a calculator to solve. You'll still find the solution using algebra, but they'll be wanting a decimal approximation for non-"nice" values, which will require "technology". An example would be:
The base of the natural logarithm is the number e (which has a value of about 2.7).
This equation has a strictly numerical term (being the 3 on the right-hand side). So, to solve this, I'll use The Relationship to convert the log equation to its corresponding exponential form, keeping in mind that the base of this log is "e":
ln(x) = 3
e3 = x
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The last value above, the cube of e, is a valid solution, and often this will be all that I'm supposed to give for the answer. However, in this case (maybe leading up to graphing or word problems) they want me to provide a decimal approximation. So I plug the expression into my calculator, and round the result on my screen. My answer is:
x = 20.086
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 before the next test.
This equation has a strictly-numerical term. So I'll be using The Relationship to convert the log equation to the corresponding exponential form. Then I'll solve the resulting equation.
log2(x) = 4.5
24.5 = x
This requires a calculator for finding the approximate decimal value. After punching a few buttons and then rounding, I find that my answer is:
x = 22.63
Solving this sort of equation usually works in this way:
If the equation has a strictly numerical term, you first use log rules to combine all log terms into one, with anything numerical on the other side of the "equals" sign. Then you use The Relationship to convert the log equation into its corresponding exponential equation, and then you may or may not meed to use your calculator to find an approximation of the exact form of the answer.
If the equation has only log terms, then you use log rules to combine the log terms to get the equation into the form "log(of something) equals log(of something else)", and then you set (something) equal to (something else), and solve.
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.
This equation has a strictly numerical term, so I'll be using The Relationship to convert the log equation to its corresponding exponential form, followed by some algebra:
log2(3x) = 4.5
24.5 = 3x
(24.5) ÷ 3 = x
x = 7.54247233266...
My final answer is:
x = 7.54
If you try to check my solution above by plugging "7.54" into your calculator for "x" in the original equation, you will get a result that is close to 4.5, but is not exactly equal. This is due to round-off error. That's not to say that you can't check your answers for log equation — you most certainly can, and probably should — but you'll need to keep this round-off-error difficulty in mind when checking your solutions. In other words, when you plug your decimal approximation into the original equation, you're just making 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, and see how close the result is to 4.5:
log2(3x) = log2(3(7.54))
At this point, I'll need to use the change-of-base formula to convert this to something my calculator knows how to deal with. I'll use the natural log:
log2(22.62) = ln(22.62) ÷ ln(2)
No, the two values are not equal, but they're pretty darned close. Allowing for round-off error, these values confirm to me that I've gotten the right answer.
If, on the other hand, my solution had returned a value of, say, 12.083, then I would have known that, no, my answer was wrong.
Expect to need to use a calculator for log-based word problems.
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Let's do a couple more examples. I picked these because they can help dispell some erroneous assumptions that people often make — because too many exercises work out the same way, so people assume that they'll always work out that way.
This equation has no strictly numerical terms; all terms are logs. So I'll be using log rules to convert each side to being just one log term. Then I'll set the arguments equal. Then I'll solve (and check!).
log4(7x + 2) + log4(9) = log4(2x)
log4((7x + 2)(9)) = log4(2x)
9(7x + 2) = 2x
63x + 18 = 2x
61x = –18
Before I say that is the answer, I first need to check (especially because this answer is negative) whether it'll work in the original equation.
But by just looking at the log term on the right-hand side of the original equation, I can see that no negative value will work.
I don't need to check the argument of the log terms on the left-hand side. I'm done. My answer is:
The example above explains why it is important to check your solutions. Sometimes, a particular solution won't work. Sometimes, no solution will work.
Don't blindly assume that every negative solution must be disallowed. Sometimes, negative solutions are valid.
This log equation has a strictly numerical term, being the 3 on the right-hand side. On the left-hand side, I have only just the one log term. So I'll be using The Relationship to convert the equation to the corresponding exponential form. Then I'll solve (and check!).
log3(x2 – 6x) = 3
33 = x2 – 6x
27 = x2 – 6x
0 = x2 – 6x – 27
0 = (x – 9)(x + 3)
x = –3, 9
Before I say that x = –3, 9 is my solution, I need to check the values (particularly that negative one). So:
log3((–3)2 – 6(–3))
= log3(9 + 18) = log3(27)
= log3(33) = 3
Huh. So the negative was okay. Go figure.
Alright; now I'll check the other solution value:
log3((9)2 – 6(9))
= log3(81 – 54) = log3(27)
= log3(33) = 3
In this case, both solutions work.
x = –3, 9