Expanding Log Expressions
Purplemath
What is "expanded form" for logarithms?
The "expanded form" of a logarithm is the form in which each log contains no multiplication or powers; at most, a log might contain a sum of terms. For example, logb(xy3), in expanded form, is logb(x) + 3logb(y).
Why do you expand logs?
We expand logs in order to rearrange them in ways that we find more useful — if expansion is indeed useful; this usefulness will vary with the log expression and the context. For instance, given log2(8x5), it might be helpful to break the variable-containing part out from the rest of the expression, as follows:
log2(8x5) = log2(8) + log2(x5)
= log2(23) + 5log2(x) = 3 + 5log2(x)
(To do the above expansion, I first applied the Product Rule, then the Power Rule, and finally the Power-of-1 Rule.)
How do you expand and simplify logs?
To expand a log expression, we apply log rules that allow us to break the log expression apart, so that we end up with each log in the expression containing no multiplication, division, or powers; and with every evaluate-able log expression having been evaluated. The idea is to make each log as plain and simple inside as possible.
However, it can be misleading to say that we are "simplifying" the original log expression, because, by splitting it apart into separate terms, we are arguably making the expression bigger, longer, and lumpier. How is that "simple"? It isn't, really.
So, at least for this topic, maybe try to ignore thoughts of "simplifying". You are expanding logs; nothing more.
- Expand log3(2x)
When the instructions say to "expand", they mean that they've given me one log expression with lots of stuff inside it, and they want me to use the log rules to take the log apart into many separate log terms, each with only one thing inside its particular log. That is, they've given me one log with a complicated argument, and they want me to convert this to many logs, each with a simple argument.
In this case, I have a "2x" inside the log. Since "2x" is multiplication, I can take this expression apart, according to the Product Rule, and turn it into an addition outside the log:
log3(2x) = log3(2) + log3(x)
There is nothing more I can do to either of these log expressions, so the answer they are looking for is:
log3(2) + log3(x)
Note: Do not try to evaluate "log3(2)" in your calculator. While you would be correct in saying that "log3(2)" is just a number (and we'll be seeing later how to rearrange this expression into something that you can evaluate in your calculator), what they're actually looking for here is the "exact" form of the log, as shown above, and not a decimal approximation from your calculator.
If you give "the answer" as being the decimal approximation, you should expect to lose points.
- Expand
I have division inside the log. According to the Quotient Rule, this can be split apart as subtraction outside the log, so:
The first term on the right-hand side of the above equation can be simplified to an exact value, by applying the basic definition of what a logarithm is. In this case, I'm using the fact that the power required on 4 to create 16 is 2; in other words, since 42 = 16, then:
log4(16) = 2
Then the original expression expands fully as:
2 − log4(x)
Always remember to take the time to check to see if any of the terms in your expansion (such as the log4(16) above) can be simplified.
- Expand log5(x3)
The exponent inside the log can be taken out front as a multiplier, so my answer is:
3log5(x)
- Expand the following:
Okay; they've told me to "expand", so I know they're wanting me to take this one log apart, into many log terms.
I'll start with the division inside this log. The 5 is divided into the 8x4, so I'll apply the Quotient Rule and split the numerator and denominator apart by converting the one log containing division into two subtracted logs:
The second term above, with just a 5 inside, is as "expanded" as that term can get, because there's only just the one thing inside the log. And, because 5 is not a power of 2, there's no further expansion that I can do. So that part of this expansion is done; I'll just be carrying the "log(5)" along for the ride to the final answer.
In the other log term, though, there's still more than just one thing inside the log. In particular, I see that there's an exponent inside the log. However, I can't take the exponent out front yet, because that power is only on the x, not the 8. I have to remember that the rule says that I can only take the exponent out front if it is on everything inside the log. So I first need to isolate that part of the argument that has the power on it.
The 8 is multiplied onto the x4, so I can split the factors inside the log by using the Product Rule to convert this one log with multiplication into two logs that are added:
log2(8x4) − log2(5)
= log2(8) + log2(x4) − log2(5)
Since 8 is a power of 2 (namely, 23), I can simplify the first log to an exact value. Because 23 = 8, then log2(8) = 3, so I get:
log2(8) + log2(x4) − log2(5)
= 3 + log2(x4) – log2(5)
Okay; now I'm finished with the first term, too; I'm only left with the middle term to expand, with the exponent inside its log.
The variable x has the exponent (which is now on everything inside its log), so I can use the Power Rule to move the exponent out in front of the log as a multiplier:
log2(8) + log2(x4) − log2(5)
= 3 + 4log2(x) − log2(5)
Each log now finally contains only one thing, and the first log term has been simplified to an exact (as opposed to approximate decimal) numerical value, so this expression is fully expanded. Then my final answer is:
3 + 4log2(x) − log2(5)
In following my work in the steps of the above computations, you may have felt that I was being a bit confusing, carrying the "log2(5)" and "3" along as I did other steps. But it is important to not drop bits of an exercise as one goes along. In whatever manner you decide to do your work — maybe including doing some of the steps, or at least portions of some of them, off to the side on scratch paper — make sure that all the steps in your final result make sense.
For instance, if I remove the talking between the steps, the previous example is worked out as follows:
= log2(8) + log2(x4) − log2(5)
= 3 + log2(x4) − log2(5)
= 3 + 4log2(x) − log2(5)
Just be careful not to try to do too many things in any one step, at least while you're just getting started on this topic.
- Expand the following:
![log_3 [ (4(x - 5)^2) / (x^4(x - 1)^3) ]](logs/logrul03.gif)
This is a gawd-awful mess! I will be taking extra care not to try to do anything in my head, nor to try to do too much all at once.
The first thing I see, inside the log, is that I've got one complicated expression that's divided by another complicated expression. To start my expansion, then, I'll split the division inside the log into subtraction of logs outside.
![log_3[{4 (x − 5)^2} / {x^4 (x − 1)^3} ] = log_3[4 (x − 5)^2] − log_3[x^4 (x − 1)^3]](logs/logrul04a.gif){kind=small}
Inside each of the two log terms I've got now, I find multiplication. So my next step will be to take apart the multiplications inside as addition of logs outside.
To make sure I don't mess up my signs, I'll be sure to put grouping symbols around the results of each split.
[log3(4) + log3((x − 5)2)]
− [log3(x4) + log3((x − 1)3)]
Now I'll take the middle "minus" through the square brackets.
log3(4) + log3((x − 5)2)
− log3(x4) − log3((x − 1)3)
Inside all but the first term, I find exponents. Since each exponent is on the entire contents of its respective log, I can go straight to moving the powers inside to being multipliers outside.
log3(4) + 2 · log3(x − 5)
− 4 · log3(x) − 3 · log3(x − 1)
Then my hand-in answer is:
log3(4) + 2log3(x − 5)
− 4log3(x) − 3log3(x − 1)
Don't think that this example is "too complicated" to show up on the next test. In fact, you should expect to see at least one question at least this complicated.
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