Algebra 2 (Eureka Math/EngageNY)
- Evaluating logarithms: change of base rule
- Logarithm change of base rule intro
- Evaluate logarithms: change of base rule
- Using the logarithm change of base rule
- Use the logarithm change of base rule
- Proof of the logarithm change of base rule
- Logarithm properties review
Using the logarithm change of base rule
Sal rewrites logarithmic expressions like 1/(logₐ4) or logₐ(16)*log₂(a) using the change of base rule.
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- How can I take this one step further to solve, say, 4^x=9?
Got it. If anyone else is curious, here’s my answer:
Okay, so I’ve got that this equals log_4 (9)=x. Then, I just have to use the change of base formula to change it to log_10 so that I can enter it in my calculator.
log_10 (4)(17 votes)
- Yes, I think you're correct. You get log_4(9) for an answer.(7 votes)
- What if the denominator has no base (e.g. log(6))? what then?(3 votes)
- What does logarithms even mean? For example log(15) is about 1.176. What does that mean? How do I use it?(3 votes)
- Usually, log means there's a hidden base of 10 (though sometimes in much higher levels of math, log means a hidden base of e). I will assume a base of 10 here.
Since the answer to a logarithm problem is an exponent, log(15) is about 1.176 means that 10^1.176 is about 15. I have confirmed this on a calculator.
Logarithms have several uses in the real world, such as the pH scale for acidity in chemistry, the Richter scale for magnitudes of earthquakes, the decibel scale for loudness of sounds, and determining how long the balance in a compound interest savings account would take to reach a certain value.
Have a blessed, wonderful day!(10 votes)
- what is the difference between log(m/n) and log m/log n?(2 votes)
- log(m/n) is the quotient or difference rule
log(m/n) = log(m) - log(n)
log(m) / log(n) is change of base rule.
log(m) / log(n) = log_n(m) . Read log base n of m.(10 votes)
- Shouldn't the 1/log_a b = log_b a be a logarithm property? If you know it as a rule it would save you quite a bunch of time when solving problems(5 votes)
- So the base of (logb)/(log4) can be anything?(4 votes)
- yes, you can use any base and log(a)/log(b) will still equal log_b(a).(2 votes)
- when do I round up to the thousand? do I round up when I solve the operation on the numerator and also on the denominator or do I keep the full numbers and round up only the final answer? I got several operations wrong because of this(3 votes)
- You round up to the thousandth at the very end of the operation. Do not round up anything whilst you are doing the expression!
It is much more better than doing the expression in one, full long line in a calculator rather than separated segments. Or you may also use the "ans" key that preserves your answer in your new calculator line.(2 votes)
- At5:26, how can Sal simply cancel out the log_10 (C)?
As log_10 (C) is in the denominator, does it matter whether log_10 (C) will equal zero or not?
Thank you :)(2 votes)
- First, he canceled the out because log(C)/log(C)=1, and anything times 1 is itself. Second, log(x), if x is anything, can't equal 0 because nothing with an exponent equals 0, unless 0 is the base, but, by default, log's base is 10(1 vote)
- I have a question, say in a test your teacher ask you to simply 1/logb(4) using the change of base rule
and you did that, but not using sal's approach, you didn't go with 1/(log4/logb). but going with 1/(log2/log2[b]). which still change of base rule.
and you get log2(b)/2, or even further you may get log2(b^2) as the answer
now my question is, (assuming I didn't make any arithmetic mistake above) will this be a "correct" answer?
consider what they asked is "using the change of base rule to simplify"(1 vote)
- I'm not sure how you made that last step.
log2(b) / 2 = log2(b) / log2(4) = log4(b). To get
log2(b^2)you'll have to arbitrarily multiply
log2(b) / 2by 4, which is not a legal operation.
Regardless, you can use any base you want, unless some particular base allows to simplify the expression further. For example, if you have an expression like
logb(c) * log(b)changing base of
logb(c)to 10 allows you to simplify the expression down to
log(c), while using some other base does not.
Also, you might be asked to give an exact answer using a calculator, and most likely it will only have
lnfunctions, so using some other base wouldn't make much sense.(4 votes)
- At around2:45, Sal does 1 ÷ (log4/logb) but if I wrote it like 1/log4/logb and then 1/log4 times 1/logb, I get 1/(log4 times logb) and not logb/log4 like Sal gets. Where did I go wrong so I don't make this mistake in the future?(1 vote)
- The parentheses matter.
Sal has the number 1 divided by the fraction (log4/logb)
You wrote the fraction (1/log4) divided by logb
Hope this helps(3 votes)
- [Voiceover] So we have two different logarithmic expressions here, one in yellow and one in this pinkish color. And what I want you to do, like always, pause the video and see if you can re-write each of these logarithmic expressions in a simpler way. And I'll give you a hint in case you haven't started yet. The hint is that if you think about how you might be able to change the base of the logarithmic, or the logarithms or the logarithmic expressions, you might be able to simplify this a good bit. And I'll give you an even further hint. When I'm talking about change of base, I'm saying that if I have the log base, and I'll color code it, log base A of B, log base A of B, this is going to be equal to log of B, log of B over log of A, over log of A. Now you might be saying wait, wait, wait, we wrote a logarithm here but you didn't write what the base is. Well this is going to be true regardless of which base you choose as long as you pick the same base. This could be base nine, base nine in either case. Now typically, people choose base 10. So 10 is the most typical one to use and that's because most peoples calculators or they might be logarithmic tables for base 10. So here you're saying the exponent that I have to raise A to to get to B is equal to the exponent I have to raise 10 to to get to B, divided by the exponent I have to raise 10 to to get to A. This is a really really useful thing to know if you are dealing with logarithms. And we prove it in another video. But now we'll see if we can apply it. So now going back to this yellow expression, this once again, is the same thing as one divided by this right over here. So let me write it that way actually. This is one divided by log base B of four. Well let's use what we just said over here to re-write it. So this is going to be equal to, this is going to be equal to one, divided by, instead of writing it log base B of four, we could write it as log of four, and if I just, if I don't write the base there we can assume that it's base 10, log of four over log of B. Now if I divide by some fraction, or some rational expression, it's the same thing as multiplying by the reciprocal. So this is going to be equal one times the reciprocal of this. Log of B over log of four, which of course is just going to be log of B over log of four, I just multiplied it by one, and so we can go in the other direction now, using this little tool we established at the beginning of the video. This is the same thing as log base four of B, log base four of B. So we have a pretty neat result that actually came out here, we didn't prove it for any values, although we have a pretty general B here. If I take the, If I take the reciprocal of a logarithmic expression, I essentially have swapped the bases. This is log base B, what exponent do I have to raise B to to get to four? And then here I have what exponent do I have to raise four to to get to B? Now it might seem a little bit magical until you actually put some tangible numbers here. Then it starts to make sense, especially relative to fractional exponents. For example, we know that four to the third power is equal to 64. So if I had log base four of 64, that's going to be equal to three. And if I were to say log base 64 of four, well now I'm going to have to raise that to the one third power. So notice, they are the reciprocal of each other. So actually not so magical after all, but it's nice to see how everything fits together. Now let's try to, now let's try to tackle this one over here. So I've log base C of 16, times log base two of C, alright. So this one, once again it might be nice to re-write these, each of these, as a rational expression dealing with log base 10. So this first one, this first one I could write this as log base 10 of 16, remember if I don't write the base you can assume it's 10, over log over log base 10 of C, and we're going to be multiplying this by, now this is going to be, we can write this as, log base 10 of C, log base 10 of C over, over log base 10 of two. Log base 10 of two. Once again I could have these little 10's here if it makes you comfortable. I could do something like that but I don't have to. And now this is interesting, cuz if I'm multiplying by log of C, and dividing by log of C, both of them base 10, well those are going to cancel out and I'm going to be left with log base 16, sorry log base 10 of 16 over, over log base 10 of two. And we know how to go the other direction here, this is going to be, this is gong to be the logarithm, log base two of 16. Log base two of 16, and we're not done yet because all this is is what power do I need to raise two to to get to 16? We'll have to raise two to the, I have to raise two to the fourth power. We did it in the blue color. To raise two to the fourth power to get to 16. So that's, this is kind of a cool thing, cuz in the beginning, I started with this variable C, it looked like we were going to have deal with a pretty abstract thing, but you can actually evaluate this kind of crazy looking expression right over here, evaluates to the number four. In fact if I had to run some type of a math scavenger hunt or something, this could be a pretty good clue for evaluating to four. You know walk this many steps forward or something. It'd be pretty cool.