Class 11 Physics (India)
Intro to logarithms
Sal explains what logarithms are and gives a few examples of finding logarithms. Created by Sal Khan.
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- is there any synonym of logarithm?(62 votes)
- Logarithm is based on the combination of two Greek words: logos and arithmos (number). Logos (λόγος) is a rather curious Greek word with multiple meanings. In this case, you could translate it as "ratio" or "proportion". The word "logarithm" was invented by John Napier in 1614.(159 votes)
- Who invented logarithms? And for what reason?(48 votes)
- Sir John Napier did... It was actually for ease of calculations, when we didn't have digital calculators. Logarithms made it easy for people to carry out otherwise difficult operations, eg: find the value of 4th root of 24. we can simply take log(24) and divide by 4. The antilog of the resultant figure will give us the answer. This is quite a feat, considering that we are not using any calculator!(67 votes)
- In what grade do you learn logarithims?
Thanks in advance.(10 votes)
- Technically, since logarithms are part of Algebra 2, by regular USA standards, probably 11th grade. Like some people have said, you can obviously learn it whenever you want, since it isn't a very complicated subject, but if you are on the average math pacing in the USA, you should learn in somewhere in 11th grade. Some people take Algebra 2 earlier than that though, so you may learn it in grades 8, 9 or 10 if you find yourself skilled and interested in math.(9 votes)
- Is log base 1 of 1 equal to 1, 0, or both?(6 votes)
- Log base 1 of 1 would have infinite answers not just 1 and 0(7 votes)
- what is zero to the zero power?(3 votes)
- That depends on whom you ask:
There are several very important mathematical theorems that require 0⁰ = 1. Thus, many or most mathematicians have simply defined 0⁰ = 1 without formal proof.
There are a few mathematicians who disagree and assert that 0⁰ is indeterminate or undefined. Particularly in certain constructions that come up in calculus (but not at this level of study) 0⁰ is a form that is indeterminate (for which there are ways to work around it and find a solution).
Thus, it really just depends on whom you ask. Most likely, your instructor would assert 0⁰ = 1, but you should ask her to see, to make sure she is not one of those who dissent from the majority of mathematicians.
NOTE: To date, no mathematician has ever been able to formally prove what, if anything, 0⁰ equals. The assertion that it equals 1 is done so that exceptions do not have to be made to those important theorems.(18 votes)
- Why x+(1/x) >=2 if x is positive and x+(1/x)<=-2 if x is negative real number?(1 vote)
- First, let's consider the case that x is positive. Note that sqrt(x) is real and nonzero since x is positive.
To show x + (1/x) >=2, it is enough to show that x + (1/x) - 2 >= 0.
The trick is to express x + (1/x) - 2 as a perfect square trinomial!
x + (1/x) - 2 = [sqrt(x)]^2 + [1/sqrt(x)]^2 - 2
= [sqrt(x)]^2 + [1/sqrt(x)]^2 - 2sqrt(x)*[1/sqrt(x)]
= [sqrt(x)]^2 - 2sqrt(x)*[1/sqrt(x)] + [1/sqrt(x)]^2
= [sqrt(x) - 1/sqrt(x)]^2
>=0 since sqrt(x) - 1/sqrt(x) is real (because x is positive).
So x + (1/x) - 2 >= 0, which then implies x + (1/x) >= 2.
Now, let's consider the case that x is negative. Then -x is positive. From the result of the first case using -x in place of x, we have
-x + [1/(-x)] >= 2 which implies -x - (1/x) >= 2 which implies -[x + (1/x)] >= 2 which implies
x + (1/x) <= -2.
Have a blessed, wonderful day!(18 votes)
- Can a logarithm be written without a base?(5 votes)
- Writing a log without a base is implied that it has base 10.
log(100) = log_10(100) = 2
The other "exception" is the natural log, which you don't write with a base.
ln(x) is the same form as log_e(x), but ln is more perferred.(7 votes)
- lets say they ask you a question that has log 100,000 and it equals 5 how do you get the answer without using the calculator(2 votes)
- There is a rule: log10^n = n
So write your problem as log 10^5, this equals 5.(9 votes)
- what is the value of n if 0.28=1/(1.09)^n? using logarithm(2 votes)
- n=log_(1/1.09)=0.28 is the logarithmic form(2 votes)
- How do you enter logarithms into a TI-83 graphing calculator?(2 votes)
- On a TI-83 Plus (hopefully it's similar for a TI-83), the logarithm key is just to the left of the 7 (it says LOG). To enter log(2), for example, you'd hit LOG, the number 2, and then close parentheses.(7 votes)
Let's learn a little bit about the wonderful world of logarithms. So we already know how to take exponents. If I were to say 2 to the fourth power, what does that mean? Well that means 2 times 2 times 2 times 2. 2 multiplied or repeatedly multiplied 4 times, and so this is going to be 2 times 2 is 4 times 2 is 8, times 2 is 16. But what if we think about things in another way. We know that we get to 16 when we raise 2 to some power but we want to know what that power is. So for example, let's say that I start with 2, and I say I'm raising it to some power, what does that power have to be to get 16? Well we just figured that out. 'X' would have to be 4. And this is what logarithms are fundamentally about, figuring out what power you have to raise to, to get another number. Now the way that we would denote this with logarithm notation is we would say, log, base-- actually let me make it a little bit more colourful. Log, base 2-- I'll do this 2 in blue... Log, base 2, of 16 is equal to what, or is equal in this case since we have the 'x' there, is equal to 'x'? This and this are completely equivalent statements. This is saying "hey well if I take 2 to some 'x' power I get 16'." This is saying, "what power do I need to raise 2 to, to get 16 and I'm going to set that to be equal to 'x'." And you'll say, "well you have to raise it to the fourth power and once again 'x' is equal to 4. So with that out of the way let's try more examples of evaluating logarithmic expressions. Let's say you had... log, base 3, of 81. What would this evaluate to? Well this is a reminder, this evaluates to the power we have to raise 3 to, to get to 81. So if you want to, you can set this to be equal to an 'x', and you can restate this equation as, 3 to the 'x' power, is equal to 81. Why is a logarithm useful? And you'll see that it has very interesting properties later on. But you didn't necessarily have to use algebra. To do it this way, to say that 'x' is the power you raise 3 to to get to 81, you had to use algebra here, while with just a straight up logarithmic expression, you didn't really have to use any algebra, we didn't have to say that it was equal to 'x', we could just say that this evaluates to the power I need to raise 3 to to get to 81. The power I need to raise 3 to to get to 81. Well what power do you have to raise 3 to to get to 81? Well let's experiment a little bit, so 3 to the first power is just 3, 3 to the second power is 9, 3 to the third power is 27, 3 to the fourth power, 27 times 3 is equal to 81. 3 to the fourth power is equal to 81. 'X' is equal to four. So we could say... Log, base 3, of 81, is equal to-- I'll do this in a different colour. Is equal to 4. Let's do several more of these examples and I really encourage you to give a shot on your own and hopefully you'll get the hang of it. So let's try a larger number, let's say we want to take log, base 6, of 216. What will this evaluate to? Well we're asking ourselves, "what power do we have to raise 6 to, to get 216?" 6 to the first power is 6, 6 to the second power is 36, 36 times 6 is 216. This is equal to 216. So this is 6 to the third power is equal to 216. So if someone says "what power do I have to raise 6 to-- this base here, to get to 216?" Well that's just going to be equal to 3. 6 to the third power is equal to 216. Let's try another one. Let's say I had, I dunno, log, base 2, of 64. So what does this evaluate to? Well once again we're asking ourselves, "well this will evaluate to the exponent that I have to raise this base 2, and you do this as a little subscript right here. The exponent that I have to raise 2 to, to get to 64." So 2 to the first power is 2, 2 to the second power is 4, 8, 16, 32, 64. So this right over here is 2 to the sixth power, is equal to 64. So when you evaluate this expression you say "what power do I have to raise 2 to, to get to 64?" Well I have to raise to the sixth power. Let's do a slightly more straightforward one, or maybe this will be less straightforward depending on how you view it. What is log, base 100, of 1? I'll let you think about that for a second. 100 is a subscript so it's, log, base 100, of 1. That's one way to think about it, I'll put parentheses around the 1. What does this evaluate to? Well this is asking ourselves, or we would evaluate this as, "what power do I have to raise 100 to, to get to 1?" So let me write this down as an equation. If I set this to be equal to 'x', this is literally saying 100, to what power, is equal to 1? Well anything that a 0 power is equal to 1. So in this case 'x' is equal to 0. So log, base 100, of 1, is going to be equal to 0. Log base anything of 1, is going to be equal to 0 because anything to the 0 power and we're not talking about 0 here. Anything that is to the power of 0 that is not 0, is going to be equal to 1.