The basic log rules are as follows:
1) log_{b}(mn) = log_{b}(m) + log_{b}(n)
2) log_{b}(^{m}/_{n}) = log_{b}(m) – log_{b}(n)
3) log_{b}(m^{n}) = n · log_{b}(m)
You should expect to need to know the basic log rules, because there is a certain type of question that the teacher can put on the test to make sure you know the rules, and know how to use the rules.
Warning: If you discover one of these trick questions on your next test, you will not be able to "cheat" your way to the answer with your calculator. If you don't know the log rules, you're toast.
Content Continues Below
Okay, let's dive into these trick problems.
The trick to doing this exercise is to notice that they've asked me to find something (namely, the log of ten) which can be created out of what they've given me (namely, the logs of two and five).
Affiliate
I just need to figure out how to apply the log rules to allow for the substitutions I'm going to need to make. So:
Since 10 = 2 × 5, and since they've given me logs of 2 and 5, then it will be helpful to me to turn the 10 into the product of 2 and 5, and then use a log rule to split the log with multiplication inside into the addition of two log expressions. In other words:
log_{b}(10) = log_{b}(2 × 5)
= log_{b}(2) + log_{b}(5)
Since they gave me values for log_{b}(2) and log_{b}(5), I can now substitute and evaluate:
log_{b}(2) + log_{b}(5)
= 0.3869 + 0.8982
= 1.2851
Then:
log_{b}(10) = 1.2851
They've asked me for the log of 9; they've given me (among other things) the log of 3. Since 9 = 3^{2}, then I can convert "9" to "3^{2}", and then apply the log rule that moves a power inside a log to being a multiplier in front of the log.
log_{b}(9) = log_{b}(3^{2})
= 2 · log_{b}(3)
Since I have the value for log_{b}(3), then I can substitutie and evaluate:
2 · log_{b}(3)
= 2 · 0.6131
= 1.2262
Then:
log_{b}(9) = 1.2262
Content Continues Below
This one is a bit more complicated, but, after fiddling with the numbers for a bit, I notice that 7.5 = 15 ÷ 2. "So what?", you ask. Well, I can create 15 from 3 and 5, and they've given me values for the logs of 3 and 5. And the division by 2 is easy to deal with, because they've given me a value for the log of 2. So:
log_{b}(7.5)
= log_{b}(15 ÷ 2)
= log_{b}(15) – log_{b}(2)
And 15 = 5 × 3, so:
log_{b}(15) – log_{b}(2)
= [log_{b}(5) + log_{b}(3)] – log_{b}(2)
= log_{b}(5) + log_{b}(3) – log_{b}(2)
And now I can substitute and evaluate:
log_{b}(5) + log_{b}(3) – log_{b}(2)
= 0.8982 + 0.6131 – 0.3869
= 1.1244
Then:
log_{b}(7.5) = 1.1244
Affiliate
Since 6 = 2 × 3, then:
log_{b}(6) = log_{b}(2 × 3)
= log_{b}(2) + log_{b}(3)
Since I have these values, I can evaluate:
log_{b}(2) + log_{b}(3)
= 0.3869 + 0.6131
= 1.0000
Then:
log_{b}(6) = 1.0000
Hmm... that was interesting. I got that log_{b}(6) = 1. Using The Relationship for logs, I get:
log_{b}(6) = 1
b^{1} = 6
b = 6
So now I know that their mysterious unnamed base "b" was actually 6. But they will not usually give you problems that let you figure out the base in this manner. In fact, they may give you questions for which the math doesn't even work out. For instance:
Doing this evaluation is easy. They've given me the logs of 3 and of 7, and 21 = 3 × 7, so:
log_{b}(21) = log_{b}(3 × 7)
= log_{b}(3) + log_{b}(7)
= 0.6 + 1.3 = 1.9
Then:
log_{b}(21) = 1.9
However, look at the other information that they included in this exercise. The value of the log of 4 should follow from the value of the log of 2, because 4 = 2^{2}. But look:
log_{b}(4) = log_{b}(2^{2})
= 2 · log_{b}(2)
= 2 · 0.4 = 0.8
Affiliate
This is not the value they gave me for log_{b}(4). They'd said that log_{b}(4) = 0.75. What's up?
Well, what's up is that they wanted to see if I really knew the log rules well enough to go straight to evaluation. Clearly, these log "values" aren't related to any actual base; the author of the exercise just picked somewhat random values. The "right" answer to the question is not the mathematically-valid one, since they weren't working with any particular log anyway; the "right" answer is the result of a correct application of the log rules, which is what this sort of question is meant to test.
If you feel at all uncomfortable with the way log rules were used in the above exercise, then you may want to re-study the rules and how they work. You can start here.
Otherwise, let's move on to the last log rule....
URL: https://www.purplemath.com/modules/logrules4.htm
© 2018 Purplemath. All right reserved. Web Design by