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 8x, so I'll apply a log rule and split the numerator and denominator apart by converting the one log with division into two subtracted logs:
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The second term above, with just a "5" inside, is as "expanded" as it 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 simplification I can do. So that part of the expansion is done; I'll just be carrying the "log(5)" along for the ride to the final answer.
Affiliate
In the first 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 x^{4}, so I can split the factors inside the log by converting to added logs:
log_{2}(8x^{4}) – log_{2}(5) = log_{2}(8) + log_{2}(x^{4}) – log_{2}(5)
Since 8 is a power of 2 (namely, 2^{3}), I can simplify the first log to an exact value. Because 2^{3} = 8, then log_{2}(8) = 3, so I get:
log_{2}(8) + log_{2}(x^{4}) – log_{2}(5)
= 3 + log_{2}(x^{4}) – log_{2}(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 a log rule and move the exponent out in front of the log as a multiplier:
log_{2}(8) + log_{2}(x^{4}) – log_{2}(5)
= 3 + 4log_{2}(x) – log_{2}(5)
Each log now finally contains only one thing, and the first log term has been simplified to a numerical value, so this expression is fully expanded. Then my final answer is:
3 + 4log_{2}(x) – log_{2}(5)
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In following my work in the steps of the above computations, you may have felt that I was being a bit confusing, carrying the "log_{2}(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:
= log_{2}(8) + log_{2}(x^{4}) – log_{2}(5)
= 3 + log_{2}(x^{4}) – log_{2}(5)
= 3 + 4log_{2}(x) – log_{2}(5)
Just be careful not to try to do too many things in any one step, at least when you're just getting started.
This is a gawd-awful mess! To do the expansion, I'll be using the log rules, and I'll be taking care not to try to do anything "in my head" or 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.
Affiliate
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.
Now I'll take the middle "minus" through the square brackets.
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.
log_{3}(4) + 2 · log_{3}(x – 5) – 4 · log_{3}(x) – 3 · log_{3}(x – 1)
Then the final answer is:
log_{3}(4) + 2log_{3}(x – 5) – 4log_{3}(x) – 3log_{3}(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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