I think it would help to clarify where the original redditor went wrong; They were trying to solve .25^x = 16, i.e. they thought the log notation meant "how do I need to exponentiate .25 to get 16" when the notation actually means "how do I exponentiate 16 to get .25". So they got the inverse operation -2 instead of the correct one, -.5

The guy was asking: isn't ¼⁻²=16? He or she doesn't understand what the logbase does. We should explain how logs work first. I.e. "what exponent of 16 gives ¼?". Then, seeing that 1/(sqrt(16)) gives ¼, you get -½ as the single exponent.

@hmmm6200

3 ай бұрын

he was saying (1/4)^-2, not positive 2 and (1/4)^-2 does equal 16 other than that u right

@Misteribel

3 ай бұрын

​@hmmm6200 thanks, I missed that. Edited to reflect it.

@jamescollier3

3 ай бұрын

he still has to teach it to all, correctly

@jamescollier3

3 ай бұрын

At the end he tapped (clicked) the board.. why didn't it erase? Is the magic broken?

@Silvar55x

3 ай бұрын

​@@jamescollier3I suspect the tap-on-board is just a distraction and the erasing is done by power of the mind alone. They probably teach the trick in calculus jedi school, or something.

The 2nd method was really cool.

That second method is quite something, loved it.

I think the main problem is that the person got the basic rule a bit backwards. Personally I changed 1/4 to 4 and put a minus sign in front of everything, then solved that easier equation.

Personally, I am not a big fan of using: a^(log_a(x) = x and then crossing out the a and the log as a standard tool for teaching - especially for high school students. I found that for the vast majority of beginners this is confusing and just promotes rote memorization rather than understanding. I much prefer to just go back to the basic definition of a log when converting to exponential form. b = log_a(c) is equivalent to c = a^b by definition. I.E. "b is the exponent we place on a to get c" The more times they repeat that phrase in their minds, the better off they will be as they keep re-enforcing what a log really is - an exponent. You did open with that definition, but, .... well anyway. For beginners - no question. Method 1.

@phiefer3

17 күн бұрын

I think I actually disagree with you on this. You say that you prefer encouraging understanding over rote memorization, but then you say that you prefer to have them repeat/memorize the definition of a logarithm as a rearranged exponent. That's literally just rote memorization though, and imho doesn't really reinforce what a logarithm really is. On the other hand, the canceling method does in fact rely on understanding what logarithms are, as well as understanding that it's an operation and the inverse of exponentiation. It also helps to move logarithms from this weird thing that you just rearrange into an exponent into a tool that can actually be used in other areas of math where sometimes your reason for taking a logarithm has nothing to do with finding exponents. ie things like using a log graph instead of the raw graph to make it easier to see certain behaviors, or in situations where the log of a function is just easier than working with the function directly. These are really strange concepts if all you think about logarithms is that they're rearranged exponents, but are much more sensible when you're used to treating logs as an operation that can be used to convert data into a more useful form.

@ianfowler9340

17 күн бұрын

@@phiefer3 You do make some good points. But when introducing logs in high school for the very first time , the definition (and understanding what that really says) is crucial to their understanding. You really think that teaching the equivalence of: 2^y = 16 and y = log_base 2(16) to newbies in high school is a weird, rote thing? It's the actual definition.I get them to rely on that equivalence to re-inforce understanding. For 16 yr olds in high school I always get them to convert: y = log_base 2(16) straight to exponential form 2^y = 16 using the definition: y is the exponent you place on 2 to get 16. If you raise both sides to the power of 2: 2^y = 2^[log_base 2(16)] , good luck on getting them to really understand why the right side is 16. You will have your work cut out for you. Having me say, "well it's the inverse" and crossing them out won't cut it for most kids in your class. To be fair, your inverse cancelling method can be used much later on when logs are second nature, but applying the basic definition to convert to exponential form will never let you down. It is certainly not weird or rote - at any level. I learned that lesson well from my 1st year university prof. Having said all of that, I do introduce, at the end of the unit, the expression: a^[log_base a(x)] and we have a good discussion on why it really does equal x . I had one student say to me:" Look sir, the the square bracket is read as ' the exponent you place on a to get x' and, well, there it is - just sitting on a." I was the 2nd happiest person in the room. Anyway, thanks for taking the time to reply. I do appreciate it.

i tried doing it with log quotient rule so log_16(1/4) = log_16(1) - log_16(4) = 0 - 1/2 = -1/2 what do u guys think?

@garrettbates2639

3 ай бұрын

This is the approach I typically use as well for these sorts of problems. It's quick and relatively painless if you have the powers of the base memorized. Though from a teaching perspective now that I look at it, I think the second way he solves it is probably more intuitive for people learning logs for the first time. I will probably be yoinking that method to use for my precalc classes from now on when we get to logarithmic equations.

Simple and direct explanation !!!

if anyone wanted to know if, the answer would be -2 if it was 1/256 instead of 1/4

5:31 cool transition

I just used the change of base formula directly. So if I have log base X of Y, but I want log base 10 of Y “Log10(Y)”, then it’s just log base 16 of X = Log10(X) divided by Log10(16) which gives -0.5 (-1/2) for X=1/4. 😃

First method is the best.

Just the trick at 1:11 on its own would have helped me better understand logarithms at school.

That didn't address the main issue is that OP confused the base and thought it was (1/4)^something instead of 16^something...

how do you do that with the markers without them slipping out of your hand, it's sorcery

@Qwentar

3 ай бұрын

It's dexterity.

@jio8464

3 ай бұрын

He has a video on how to manipulate the markers

Briefly said: 16 is the base - so the mistake is inside the question.

Haven't seen this neat formula in method two before. (Or maybe I did see it and forgot.) I used a related approach, the change of formula but I used the binary logarithm because both ¼ and 16 can easily be written as powers of 2: log₁₆(¼) = log₂(¼) / log₂(16) = log₂(2⁻²) / log₂(2⁴) = -2/4 = -½ Done! 😊

Round about way of telling them that 1/4 wasn't the base of the power they are suppose to be taking but 16 was the base.

Can u explain about the logarithms being the inverse of exponentiatio. Please. I need to understand step by step how raising that 16 to that logarithmic cancels the log part. It would help me a lot actually coz I'm trying to understand stuff so i know why i should do this and not that ... Thanks :)

Making each side the exponent to the same base doesn't seem like it would always be equal.

@Krisztian1941

3 ай бұрын

It must be equal because the function y=4^x is always strictly increasing. So no 2 different x's will give the same y as a result. That's why you can just cross out the base and continue with the exponents as a linear equation.

@michaeljordan215

3 ай бұрын

@@Krisztian1941 what I meant was as a general application to all equations. It doesn't seem to me that you can make both sides of the equation the exponent of a base and it would remain equal. Not like adding the same number to both sides always keeps the equation true.

@phiefer3

3 ай бұрын

@@michaeljordan215No, it must always be equal when you do that. If x=y then for any n, n^x must be equal to n^y because if the bases are equal (ie they're both n) and the exponents are equal (because x=y), then the results must be equal as you are doing the same thing on both sides. If doing this ever wasn't true then the starting equation must have been false to start with.

@blizzaga1

3 ай бұрын

​@@michaeljordan215 If bases are equal then exponents must also be equal unless you have 0 or 1 as a base.

@mouykaing7456

3 ай бұрын

@@michaeljordan215 No, this is always true because if b^x=b^y, you can always take the log base b on both sides to get x=y, unless b equals 0 or 1. This is like saying that if x/a=y/a, it's not always the case that x=y. If these things weren't always the case (except for some special values), algebra as we know it wouldn't work.

I am a 7th grader and I solved it in first try💀

@xtranub8792

2 ай бұрын

I never knew they teach log in 7th grade I myself learnt it in 11th

Super

Second

First

first

@arnoygayen1984

3 ай бұрын

I hereby officially declare you as first.