I am 63 years old and never did well in math. I work around a bunch of engineers and am truly amazed with math at this point in my life. I love your videos. I like the way you explain simple math. I am at this level, so I really appreciate it.
I find your comment very interesting. If I'm not being too personal, I am really curious to learn how you came to be working around a bunch of engineers without having done well with math? 🙂
@@cheriem432 In High School I worked a full day so my math was limited. I never went on to calculus and really didn't do great at algebra. I am now a construction manager for renewable power plants and work closely with all types of engineers.
Interesting use of the phrase "round it off" there. log9 ÷ log2 = 3.169925... Rounding that off to three decimal places, we get 3.17. (The third place would be a nought, which we can omit.) 3.169 is a truncation, not a rounding.
@@dazartingstall6680 I think its more to avoid the arguments of when to use either method at the end of the vid as its a whole subject in itself . its a good point though !
Interesting.. not had to even think of logs for decades so a good revision for me. The phrase "round it off" there is very wrong. Using four figure log (tables) you can only quote the answer to three significant figures anyway but suppose you used six figure or were allowed to use a calculator then the question should either state or you should state as part of your answer the number of significant figures. So if you used a calculator say you might wish to quote to four figures giving us 3.170 and here the zero is a significant number. Extending the solution a little an astute student might point out that the rounding error is so infinitesimally small that for practical purposes the solution is 3.17
Once the definition of logarithm is shown at 6:30 what is there to show? 2^x=9 can be rewritten as log base 2 of 9 =x by definition. Then you can get the answer from the Google search bar QED No need to take the log of both sides and use a property or two. Then you can go into the story about how in the old days mathematians spent 20+ years calculating tables of logarithms.
It took a little while to integrate this material into my understanding of mathematics, but with the help of this utube link I have grasped the concept. Thanks.
I am 81 and I need to go find a slide rule. Anyway, I think you gave a great explanation. I looked at this in an effort to ward off dementia. I hope it works. You did a great job and keep up the work.
2^x = 9 off top x=1 2^1 = 2 2^2 = 4 2^3 = 8 2^4 = 16 guess x = 3.2 for 2^x=9 this is an excellent problem for logs. log a = b or 10^b = a further loga^x = xloga [basic log rule] log2 = x where 10^x = 2 log2 = 0.3010 log9 = 0.9542 [found on calculator but back in my day we looked up these numbers in the back of our text book in log tables because we had no calculators. We even had logs of trig functions. The little black book for the engineer exam has (or ¿had?) these tables.] 10^0.3010=1.9999, =2 10^0.9542=8.9991, =9 here 2^x = 9 same on either side=> log { 2^x = 9 } log(2^x) = log(9) xlog2= log9 x= log9/log2 = 0.9542/0.3010 = 3.1699 our previous guess = 3.2 V E R I F Y (calculator) 2^3.1699 =? 9 8.9998 =❤ 9✔️
Further verification, again with a calculator: 2^(log9 ÷ log2) = 9 I also remember log tables. We had them in a thin book that also contained sine, cosine and tangent tables.
You just committed an error with your calculator because you wrote .9542/.3010 = 3.169925 . You really used log(9)/log(2) , but wrote the truncated intermediate results of each log as you typed in the expression, and then cited more precision than what you wrote for the truncated log values. [With the full precision of the calculator, x ≈ 3.1699250014423...] If you actually entered .9542 and.3010 , you'll get 3.170{0996677740...} . This is good for 4 sig figs, 3.170 , because the inputs are only good for 4 sig figs. I simplified my solution using log(3): 2ˣ = 9 x = log(9)/log(2) = 2*log(3)/log(2) ≈ 2*0.47712/0.30103 ≈ 3.1699 I remember the log of 2 and 3 to many sig figs, and I used 5 places because their values drop off nicely at 0.30103 and 0.47712 . A simple 4-function calculator returns 3.1699{166196060...} , so the estimates for the logs yield 5 sig figs with 3.1699 .
Without a calculator as sort of an approximation, I took 2+2+2+1 = 9, where 1 is 1/2 of 2 and split 3 ways is 1/3, so 1/3 of 1/2 is 1/6 or .16666 or .17 rounded to the hundreds, which is kinda close.
@@tomtke7351 It also used subtraction. I remember you had to mentally move the decimal point for bigger numbers. So you kind of needed the language of maths to use the higher functions. Maybe I'll dig it out and see how far I can still use it. My Dad had a circular one in a case.
thanks good person. 1 tiny problem - it was a repeating decimal. but you are right. many problems - have good , simple , but excellent answers. in other words - try harder. keep the Faith. Math Good , amen.
The next Great Leap Forward in Mathematics should be to convert everything to Base Twelve! We already use twelves because of all the even factors of 2, 3, 4 and 6, but we clumsily convert them into tens since we have five fingers on each hand. However, each of our hands has four fingers each with three segments, so counting by twelves should be easy! Besides that, each year consists of approximately twelve months (or lunations) which we divide into four seasons of three months each, as if the universe is trying to tell us something. Counting by twelves makes it easy to divide 2, 3, 4 and 6 evenly so we're actually already doing it, just more clumsily by using tens as a base. The British have actually done a lot of work on this system so it's already to be adopted whenever enough people get over their obsession with tens ;)
This makes no sense! The tens, hundreds, thousands etc notation is far, far easier than counting on your fingers. Which is why the metric system is so superior to anything else a couple of countries are still stupidly holding on to.
I have forgotten just about everything I ever learned about exponents and logarithms. I don't think I ever understood logs, anyway. You confused me by going in depth about the " bacon and eggs" leading me to see log base 2 of 9 equals x. But then you solved it by ignoring that and turning it into x times log (base 10) of 2 equals log9. So the REAL lesson for me was that log 2 to the x equals x times log 2. That was what I needed to know and you kind of just brushed that aside. Nevertheless, I (re)learned something, so thanks.
This is a serious question. How many people NEED to know how to solve this equation? This leads to a further consideration and that is : In High School Maths already a lot of topics are included that 90% of pupils will NEVER need nor understand. This can mean that many pupils simply give up on ALL of Math thus leaving them struggling with daily calculations that are useful, and at risk of being cheated by people who can do basic math. I am not suggesting that a full math program should not be offered to those who have interest and ability. I have a daughter who is professor of astrophysics, but that is unusual. So she benefited from a comprehensive math course at school. On the other hand I recently looked at a Year 10 math course from England most of which I simply did not recognise. I concede that if I had understood calculus in 1961 my whole life would have been different, but not necessarily better.
@@mikesheth5370 I assumed 9/2 was how you'd arrived at 4.5. We find the base 10 log of 9 We find the base 10 log of 2 We divide the former by the latter The number 4.5 does't appear anywhere in this calculation.
Thanks for teaching this again at my age of over half a century. When I was going to school back then too young too understand this things. Like putting a kid in a rocket classroom.
Mathematicians use the words "logarithm" and "exponential" far too carelessly. The log function takes an ordinary number and moves it into the logarithmic number domain. In the case log 10 =1 the 10 is a flat finger counting number and the 1 is the logarithm. But the logarithmic number is the same as the exponential number (base 10). Therefore the 16th century function name "logarithm" is redundant, we should rename it as, or just describe it as, the exponential transform function. This is important because we stop having two names for the 1 in my example. At present I can describe the 1 as a logarithmic number or as an exponential number but the 1 does not change, it is still the same 1 in the reverse exponential evaluation of the 1 i.e. 10^1 = 10. The point is that e^x should not be called the exponential function, it is the reverse exponential transform; the x is the natural exponential number and e^x decrypts the x into a flat finger counting number.
This video is for students of fundamental algebra. Your comment, while interesting, belongs more on a blog of number theory. You wouldn’t walk into a first grade classroom, where they teach 1+1=2 and start explaining that without the additive property, mathematics has no concrete foundation. Their eyes would glaze over and they’d be shoving cookies in their mouths.
Greetings. The identity can be solved by the use of log. Therefore, 2^X=9, 2^X=3^2, an Log2^x=Log3^2, XLog2=2Log3, and X=2Log3/Log2, X=2(1.58496)= 3.16993, X approximately equals 3.17.
You're sloppy with your notation. x = 2*log(3)/log(2) x ≈ 2*1.58496 ≈ 3.16992 You shouldn't use "=" for an approximation. And once you cite a certain value for an approximation, you can't use another value with more precision. That is, the last digit is '2', not '3'. It's obvious you copied a few digits from your calculator, but continued the calculation with more precision. If you had cited x ≈ 2*1.584962500... , then you can show more digits than what you had typed.
The fun thing about getting older is that I can watch a movie and it is much like watching it for the first time. 😊. Same with Maths. I unused to be a Maths tutor in University and at a private tutorial center. I cat believe how much I have forgotten. Thank you.
@@TheSimCaptain To convert Base 2 to base 10, using his B, A and E notation: '2^x=9' => 'log(base 2) 9 = x' => 'log A/log B = E' where 'A = 9', 'B = 2' and 'E' is the exponent and the value of 'x'/ :)
I'm happy to say I remembered that Logs would be involved but in spite of remember doing this stuff I"d forgot the 2^x to x log 2 rewrite. How old you might ask. Well, after you showed how to do it I took out my slide rule and verified it. (Yes, I have one of those things on my computer desk. And I don't have to worrying about remembering to change the batteries 🤣)
1 of my high school 🏫 STEM teachers (double major BS in Math and Physics a major liberal arts college in the East Coast) would give 1/2 credit. Why? He would point out that 9 = 3^2. Thus, (2 ln 3)/ ln 2 = x or 2 [(ln 3) / (ln 2)] as the final answer.
@@WineSippingCowboy true, mine is a general solution whereas your solution only works if the right-hand side has a rational square root. How would you solve it if the right-hand side is, say, 11?
I must admit I do know this one has to be solved with log but I don't have the routine to get the equation right. Acoustics wasn't my thing at school. Edit: I couldn't stand I had forgotten these terrible log equations so I forced myself to learn this AND solving polynomials after dealing with 8^y + 2^y = 130 or (2^y)^3 + (2^y) = 130 with x = 2^y so x³ + x = 130 So first there is this 3th grade equation x³ + x -130 = 0 that I couldn't solve. But I learned to give a try with some factors of -130 being -/+ 1, 2, 5, 10 f(1) = 1³ + 1 - 130 neh f(2) = 2³ + 2 - 130 neh f(5) = 5³ + 5 - 130 = 0 BINGO so x = 5 now we have to find the quadratic equation (x - 5) . (x² + bx +26) 1 . x³ + 0 . x² + 1 . x - 130 x 5 5 25 130 -------------------------------------------------- + 1 5 26 0 so x² + 5x + 26 = 0 or (x - 5) . (x² + 5x + 26) = 0 D = b² - 4ac = 25 - 4 . 1 .26 = 25 - 104 = - 79 so no further rational solutions. x = 2^y = 5 and then the logs come in: y . log 2 = log 5 so y = log 5 / log 2 ~ 2.32 units
This is lovely. I love maths so much but unfortunately didn't reach this level of education. I would like to access your lessons to advance my knowledge and thanks for this ❤❤❤.
Please, for us (dummies) unfamiliar with these procedures can you explain why x squared gets added to one, and what happens with dx? A lot of the time it's these (perhaps insignificant) details left unexplained that leave many of us puzzled as to why a certain action is taken. Thank you.
@@mauriziograndi1750: I have a text file filled with a bunch of math symbols, Greek letters, subscripts, superscripts, etc. for the purpose of typing out my math comments. I use Copy & Paste when I need a symbol that isn't available on my keyboard. I also use standard notation that's accepted by math software, programming languages, and software tools. Thus, I prefer to write ln(9)/ln(2) or log(9)/log(2) because Ln9 and ln2 are variables, and not ln(9) and ln(2), unless assigned to those values.
It’s been years since I’ve done math like this. I remember looking up logarithms on the chart in out textbook. We didn’t have the cool calculators at that time.
I looked up my Chambers seven decimal place log tables and found that log 9 (to base 10) was 0.9542 (to four decimal places) and log2 was 0.3010, so 0.9542/0.3010 = approx, 3.17 for x.
^ --> means raise to the power of 2^x = 9 --> when searching for the exponent, logarithms are involved Rewrite 2^x = 9 to logarithmic form log₂(9) = x log₂(9) = x 2log₂(3) = x ------------------------------------------------------- log₂(9) = x log₂(3 * 3) = log₂(3) + log₂(3) = x ------------------------------------------------------- log₂(9) = x (log(9)) / (log(2)) = 3 ---> the base-2 log change to natural log
This yt uploader is probably one of them. Even if the solution is, somehow, magically correct, the delivery style does not teach an objective lesson unobstructed.
Even someone that took algebra in high school likely never heard of natural or base 10 logs. So 90% of people would have no idea. And it wouldn't effect their lives, effect the jobs they would chose to do and want to do.
I was really hoping this video would actually talk about how to calculate a Log; that's one area of algebra that's always been a mystery to me. I get that Log 9 is a value, but HOW do you calculate it? I was hopeful until you said to take out your calculator to find the final answer. I think that would go along way to helping folks understand Logs, if we knew what the calculator was actually doing instead of it being a magic box and spitting out some never-ending decimal number. I've just never been able to visualize what its doing. Thank you.
Sorry, there is no simple calculation. Honest. To derive the formula used to calculate log(x), you need to use calculus. The basic calculation involves summing an infinite series - in practice keep adding successive terms until you have the accuracy you require. That's why most people use a calculator or a Big book of pre-calculated values of log(x). here is one formula for the natural log of x: ln(x) = sum (for k=1 to infinity) [ (1/k) * ((x-1)/x)^k ] valid for x >= ½ to get 15 decimal places of accuracy for ln(9) you need to evaluate about 250 terms of the form (1/k)*(8/9)^k . for k=1 to k=250 so ln(9) ~ 2.19722457733619 You need more digits, add up more terms.
@@Steven-v6l I shared that same interest in learning how to find the log; so thanks for sharing your knowledge in the reply. Your post makes a great argument for investing in a calculator!!
Ok. What? I get 2 x 2 x 2 = 8, i.e. 2 raised to the power 3. I get that! And I get 2 raised to the power 4 is 16. I get that too! But what the heck is 2 raised to the power 3.169?? 2 x 2 x 2 x 0.169? Which is, of course, wrong. I realise that 0.169 is spread across each 2! So, dividing 0.169 by 3 (for each 2) and adding that to 2 it's very roughly something like (but isn't quite) 2.056333333 x 2.056333333 x 2.056333333 which sort of equals 9. I now understand what 2 raised to the power of 3.169 kind of is. But now you realise what you're dealing with here 😆
Without a calculator, and per your first explanation wouldn't bet best answer be log2 (9) = x? I agree that using log10 (or ln) makes sense when using a calculator, but it adds a step of taking the log of both sides then doing the division, where as you can just use the identity.
I'm adding as a new comment as well as a reply elsewhere as this is important: Interesting.. not had to even think of logs for decades so a good revision for me. The phrase "round it off" there is very wrong. Using four figure log (tables) you can only quote the answer to three significant figures anyway but suppose you used six figure or were allowed to use a calculator then the question should either state or you should state as part of your answer the number of significant figures. So if you used a calculator say you might wish to quote to four figures giving us 3.170 and here the zero is a significant number. Extending the solution a little an astute student might point out that the rounding error is so infinitesimally small that for practical purposes the solution is 3.17
As a question, when I do lg9/log2 in my calculator, I get the value 3.16992500144 within the limits of my display. Wouldn't the 3 orders of magnitude precision give 3.170 instead of 3.169, as the next 9 is greater than 5, or is this a situation where because the number that one would round up from is an odd number, you don't round up? (At this level of precision, there isn't a hand tool you would use that would recognize the difference, though a machinist would probably complain.)
Why isn't this the equation: Log (base2)9=X? I ask this based on the relationship you show at the 6:05 mark. I am just following your rule for 2(exp)X=9.
@@victorstalick5528: No antilog is needed. The solution is x = log(9)/log(2) = 2*log(3)/log(2) . We can approximate it using 0.47712 and 0.30103 for log(3) and log(2) respectively.
I only do the so called 4 simple maths. However I'd make a guess 4.5, now I will watch the video and see where I an wrong.. I am an old man in my 70's and I never needed this kind of math.
I am 63 years old and never did well in math. I work around a bunch of engineers and am truly amazed with math at this point in my life. I love your videos. I like the way you explain simple math. I am at this level, so I really appreciate it.
I find your comment very interesting. If I'm not being too personal, I am really curious to learn how you came to be working around a bunch of engineers without having done well with math? 🙂
@@cheriem432 In High School I worked a full day so my math was limited. I never went on to calculus and really didn't do great at algebra. I am now a construction manager for renewable power plants and work closely with all types of engineers.
Same for me. I am 52 years old and my best friends are engineers. Needless to say, I try not to discuss mathematics when they are around!
Interesting use of the phrase "round it off" there.
log9 ÷ log2 = 3.169925...
Rounding that off to three decimal places, we get 3.17. (The third place would be a nought, which we can omit.)
3.169 is a truncation, not a rounding.
it's correct
@@jenohathazi920 I didn't say it's not correct. But it's a correct truncation, not, as John called it, a rounding off.
@@dazartingstall6680 I think its more to avoid the arguments of when to use either method at the end of the vid as its a whole subject in itself . its a good point though !
Interesting.. not had to even think of logs for decades so a good revision for me. The phrase "round it off" there is very wrong. Using four figure log (tables) you can only quote the answer to three significant figures anyway but suppose you used six figure or were allowed to use a calculator then the question should either state or you should state as part of your answer the number of significant figures. So if you used a calculator say you might wish to quote to four figures giving us 3.170 and here the zero is a significant number. Extending the solution a little an astute student might point out that the rounding error is so infinitesimally small that for practical purposes the solution is 3.17
The value to 3 decimal places is 3.170 . The way you wrote your value indicates 2 decimal places.
2^x=9
exponential writing
e^(x ln(2))=e^ln(9)
x ln(2)=ln(9)
x=ln(9)/ln(2)
Thank you it's a pleasure to ear your explanations
Overcomplicated! X = log9(base2) by definition of logarithm
Once the definition of logarithm is shown at 6:30 what is there to show?
2^x=9 can be rewritten as log base 2 of 9 =x by definition. Then you can get the answer from the Google search bar QED
No need to take the log of both sides and use a property or two. Then you can go into the story about how in the old days mathematians spent 20+ years calculating tables of logarithms.
It took a little while to integrate this material into my understanding of mathematics, but with the help of this utube link I have grasped the concept. Thanks.
Could you please differentiate the proof?
This is one way of giving people headaches, but very cool. Thank you.
I am 81 and I need to go find a slide rule. Anyway, I think you gave a great explanation. I looked at this in an effort to ward off dementia. I hope it works. You did a great job and keep up the work.
The log of "a" to the base "b", is equal to the log(a), divided by the log(b). Here, a = 9, and b = 2. Hence: x = log(9) / log(2)
2^x = 9
off top
x=1
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
guess x = 3.2 for 2^x=9
this is an excellent problem for logs.
log a = b
or
10^b = a
further
loga^x = xloga [basic log rule]
log2 = x where 10^x = 2
log2 = 0.3010
log9 = 0.9542
[found on calculator but back in my day we looked up these numbers in the back of our text book in log tables because we had no calculators. We even had logs of trig functions. The little black book for the engineer exam has (or ¿had?) these tables.]
10^0.3010=1.9999, =2
10^0.9542=8.9991, =9
here
2^x = 9
same on either side=>
log { 2^x = 9 }
log(2^x) = log(9)
xlog2= log9
x= log9/log2
= 0.9542/0.3010
= 3.1699
our previous guess = 3.2
V E R I F Y (calculator)
2^3.1699 =? 9
8.9998 =❤ 9✔️
Further verification, again with a calculator:
2^(log9 ÷ log2) = 9
I also remember log tables. We had them in a thin book that also contained sine, cosine and tangent tables.
You just committed an error with your calculator because you wrote .9542/.3010 = 3.169925 . You really used log(9)/log(2) , but wrote the truncated intermediate results of each log as you typed in the expression, and then cited more precision than what you wrote for the truncated log values. [With the full precision of the calculator, x ≈ 3.1699250014423...]
If you actually entered .9542 and.3010 , you'll get 3.170{0996677740...} . This is good for 4 sig figs, 3.170 , because the inputs are only good for 4 sig figs.
I simplified my solution using log(3):
2ˣ = 9
x = log(9)/log(2) = 2*log(3)/log(2)
≈ 2*0.47712/0.30103 ≈ 3.1699
I remember the log of 2 and 3 to many sig figs, and I used 5 places because their values drop off nicely at 0.30103 and 0.47712 . A simple 4-function calculator returns 3.1699{166196060...} , so the estimates for the logs yield 5 sig figs with 3.1699 .
I love how you talk in circle so you can generate and log more minutes to RUclips 😅!
Without a calculator as sort of an approximation, I took 2+2+2+1 = 9, where 1 is 1/2 of 2 and split 3 ways is 1/3, so 1/3 of 1/2 is 1/6 or .16666 or .17 rounded to the hundreds, which is kinda close.
I understood from basic knowledge of exponents that it would be 3-something. Thanks for explaining what a logarithm is!😊
It's for beginners / those who know first principals only.. I a BSc student of 1970s enjoyed the way..
Thankyou for taking the time to fully explain the options to be considered.
I remember this from when we were taught about logs and slide rules! - Early 1960s. Thank you for stirring the memory.
add in: the slide rule is logarithmic laid out to essentislly ADD together two numbers' logs when multiplying them.
@@tomtke7351 It also used subtraction. I remember you had to mentally move the decimal point for bigger numbers. So you kind of needed the language of maths to use the higher functions. Maybe I'll dig it out and see how far I can still use it. My Dad had a circular one in a case.
❤❤❤.
thanks good person. 1 tiny problem - it was a repeating decimal. but you are right. many problems - have good , simple , but excellent answers. in other words - try harder. keep the Faith. Math Good , amen.
The next Great Leap Forward in Mathematics should be to convert everything to Base Twelve! We already use twelves because of all the even factors of 2, 3, 4 and 6, but we clumsily convert them into tens since we have five fingers on each hand.
However, each of our hands has four fingers each with three segments, so counting by twelves should be easy!
Besides that, each year consists of approximately twelve months (or lunations) which we divide into four seasons of three months each, as if the universe is trying to tell us something. Counting by twelves makes it easy to divide 2, 3, 4 and 6 evenly so we're actually already doing it, just more clumsily by using tens as a base.
The British have actually done a lot of work on this system so it's already to be adopted whenever enough people get over their obsession with tens ;)
This makes no sense! The tens, hundreds, thousands etc notation is far, far easier than counting on your fingers. Which is why the metric system is so superior to anything else a couple of countries are still stupidly holding on to.
thanks for the lesson
An experienced teacher.Thak you so much for making maths much easier.
I have forgotten just about everything I ever learned about exponents and logarithms. I don't think I ever understood logs, anyway. You confused me by going in depth about the " bacon and eggs" leading me to see log base 2 of 9 equals x. But then you solved it by ignoring that and turning it into x times log (base 10) of 2 equals log9. So the REAL lesson for me was that log 2 to the x equals x times log 2. That was what I needed to know and you kind of just brushed that aside. Nevertheless, I (re)learned something, so thanks.
Great video it’s so easy to forget this stuff you make it easy to remember & understand
Detailed explanation made it more understanding for all knowing why they do what they do.
2ˣ = 9
x = log(9)/log(2) = 2*log(3)/log(2)
≈ 2*0.47712/0.30103 ≈ 3.1699
This is a serious question. How many people NEED to know how to solve this equation? This leads to a further consideration and that is : In High School Maths already a lot of topics are included that 90% of pupils will NEVER need nor understand. This can mean that many pupils simply give up on ALL of Math thus leaving them struggling with daily calculations that are useful, and at risk of being cheated by people who can do basic math.
I am not suggesting that a full math program should not be offered to those who have interest and ability. I have a daughter who is professor of astrophysics, but that is unusual. So she benefited from a comprehensive math course at school.
On the other hand I recently looked at a Year 10 math course from England most of which I simply did not recognise.
I concede that if I had understood calculus in 1961 my whole life would have been different, but not necessarily better.
2ˣ = 9
log2ˣ = log9
x(log2) = log9
x = log9/log2 (I hope)
Yes simple if you know Logarithme solution. Log on base 10. Look up tables and solved.Log on base 10 for 4.5.
@@mikesheth5370 log9/log2 isn't the same as log(9/2).
This is also equal to log base 2 of 9.
@@dazartingstall6680 Have to check like g theorems! Can find log value on base 10 for 9 and 2 and divide.
@@mikesheth5370 I assumed 9/2 was how you'd arrived at 4.5.
We find the base 10 log of 9
We find the base 10 log of 2
We divide the former by the latter
The number 4.5 does't appear anywhere in this calculation.
Where to start depends on what you want. In short, X is by definition log 9(base2)
Then he says common Log base 10, after a lengthy waffle on side-track.
We used Clarks Logarithmic table at school and in college,it was slide rules. Really enjoyed reminiscing.
You are a good teacher in Math
i appreciate your kind effort in doing math explanation.
Take log on both sides we get x=log9 to base 2
Thanks for teaching this again at my age of over half a century.
When I was going to school back then too young too understand this things.
Like putting a kid in a rocket classroom.
Mathematicians use the words "logarithm" and "exponential" far too carelessly. The log function takes an ordinary number and moves it into the logarithmic number domain. In the case log 10 =1 the 10 is a flat finger counting number and the 1 is the logarithm. But the logarithmic number is the same as the exponential number (base 10). Therefore the 16th century function name "logarithm" is redundant, we should rename it as, or just describe it as, the exponential transform function. This is important because we stop having two names for the 1 in my example. At present I can describe the 1 as a logarithmic number or as an exponential number but the 1 does not change, it is still the same 1 in the reverse exponential evaluation of the 1 i.e. 10^1 = 10. The point is that e^x should not be called the exponential function, it is the reverse exponential transform; the x is the natural exponential number and e^x decrypts the x into a flat finger counting number.
This video is for students of fundamental algebra. Your comment, while interesting, belongs more on a blog of number theory. You wouldn’t walk into a first grade classroom, where they teach 1+1=2 and start explaining that without the additive property, mathematics has no concrete foundation. Their eyes would glaze over and they’d be shoving cookies in their mouths.
Me and math are polar opposites and don't attract. I learned my times tables up to 12x12. And that served me very well for 60 years.
skip to 10:04 and you’re good
Greetings. The identity can be solved by the use of log. Therefore,
2^X=9,
2^X=3^2, an
Log2^x=Log3^2,
XLog2=2Log3, and
X=2Log3/Log2,
X=2(1.58496)=
3.16993, X approximately equals 3.17.
You're sloppy with your notation.
x = 2*log(3)/log(2)
x ≈ 2*1.58496
≈ 3.16992
You shouldn't use "=" for an approximation. And once you cite a certain value for an approximation, you can't use another value with more precision. That is, the last digit is '2', not '3'. It's obvious you copied a few digits from your calculator, but continued the calculation with more precision. If you had cited x ≈ 2*1.584962500... , then you can show more digits than what you had typed.
This would've been so helpful with my online intro to electrical engineering class
The fun thing about getting older is that I can watch a movie and it is much like watching it for the first time. 😊. Same with Maths. I unused to be a Maths tutor in University and at a private tutorial center. I cat believe how much I have forgotten. Thank you.
So how do you calculate 2 to the 3.167 power?
Many thanks
Blessing to you always
Thank you!
If you took log to the base of 2, then the LHS would resolve to x. Then x=log base 2 of 9
This is exactly my answer.
The only problem is you won't find log to the base 2 in your calculator. You're still left with the same problem of finding the arithmetic value.
@@TheSimCaptain that's funny, I can do it on mine - it allows choosing your base.
@@jamesharmon4994 Not on most calculators. But you still need a calculator, as log base 2 of 9 isn't an arithmetic solution.
@@TheSimCaptain To convert Base 2 to base 10, using his B, A and E notation: '2^x=9' => 'log(base 2) 9 = x' => 'log A/log B = E' where 'A = 9', 'B = 2' and 'E' is the exponent and the value of 'x'/ :)
I'm happy to say I remembered that Logs would be involved but in spite of remember doing this stuff I"d forgot the 2^x to x log 2 rewrite. How old you might ask. Well, after you showed how to do it I took out my slide rule and verified it. (Yes, I have one of those things on my computer desk. And I don't have to worrying about remembering to change the batteries 🤣)
This is how I would do it:
2^x = 9
ln(2^x) = ln(9)
xln(2) = ln(9)
x = ln(9)/ln(2)
1 of my high school 🏫 STEM teachers (double major BS in Math and Physics a major liberal arts college in the East Coast) would give 1/2 credit. Why? He would point out that 9 = 3^2. Thus, (2 ln 3)/ ln 2 = x or 2 [(ln 3) / (ln 2)] as the final answer.
@@WineSippingCowboy true, mine is a general solution whereas your solution only works if the right-hand side has a rational square root. How would you solve it if the right-hand side is, say, 11?
@@paulkurilecz4209 Your solution would be the model. 11 in place of 9.
I must admit I do know this one has to be solved with log but I don't have the routine to get the equation right. Acoustics wasn't my thing at school.
Edit: I couldn't stand I had forgotten these terrible log equations so I forced myself to learn this AND solving polynomials after dealing with 8^y + 2^y = 130 or (2^y)^3 + (2^y) = 130 with x = 2^y so x³ + x = 130
So first there is this 3th grade equation x³ + x -130 = 0 that I couldn't solve. But I learned to give a try with some factors of -130
being -/+ 1, 2, 5, 10
f(1) = 1³ + 1 - 130 neh
f(2) = 2³ + 2 - 130 neh
f(5) = 5³ + 5 - 130 = 0 BINGO so x = 5 now we have to find the quadratic equation (x - 5) . (x² + bx +26)
1 . x³ + 0 . x² + 1 . x - 130
x 5 5 25 130
-------------------------------------------------- +
1 5 26 0
so x² + 5x + 26 = 0 or (x - 5) . (x² + 5x + 26) = 0 D = b² - 4ac = 25 - 4 . 1 .26 = 25 - 104 = - 79
so no further rational solutions.
x = 2^y = 5 and then the logs come in: y . log 2 = log 5 so y = log 5 / log 2 ~ 2.32 units
This is lovely. I love maths so much but unfortunately didn't reach this level of education. I would like to access your lessons to advance my knowledge and thanks for this ❤❤❤.
What if we don't have scientific calculator? I remember referring to anti-log tables to calculate log of a number to base 10 .
Please, for us (dummies) unfamiliar with these procedures can you explain why x squared gets added to one, and what happens with dx? A lot of the time it's these (perhaps insignificant) details left unexplained that leave many of us puzzled as to why a certain action is taken. Thank you.
Ln9 / ln2 = 3.169925
2^3.169925 = 9
"≈" yes; "=" no.
@@oahuhawaii2141
You right
@@mauriziograndi1750: I have a text file filled with a bunch of math symbols, Greek letters, subscripts, superscripts, etc. for the purpose of typing out my math comments. I use Copy & Paste when I need a symbol that isn't available on my keyboard.
I also use standard notation that's accepted by math software, programming languages, and software tools. Thus, I prefer to write ln(9)/ln(2) or log(9)/log(2) because Ln9 and ln2 are variables, and not ln(9) and ln(2), unless assigned to those values.
thanks for being and working ,trying to be a great Teacher. Good work.
I never got an explanation of Log before. Thank you
2^x = 9 = 3^2
x = (2log3)/(log2)
I just rewrote the original equation as a log equation. log base 2 (9) = x. Which can be rewritten using rules of logs as Log 9/Log 2.
Simple! If you know log solution . Log of both side. and power can be multiplied X Log2 = Log8
The morning after , this is just what I was looking for to get me through . We all could use some continuing education right about now .
Just the same. This and cooking videos. And possibly a long trip somewhere....
It’s been years since I’ve done math like this. I remember looking up logarithms on the chart in out textbook. We didn’t have the cool calculators at that time.
What about doing the problem without the calculator?
Use either a log table or a slide rule.
Those are also calculators. He means show your worked out solution using pen and paper. Did you even apply math?
Take the natural log of both sides and one has xln2 =ln9. This is really xln2 = 2ln3. x = 2ln3/ln2.
x ~ 3.1699
I love the reminders of 55 years ago, but I would like the presentation a little more direct. Whatever I will keep watching.
Good video. Thanks for producing.
Just as a side, you could graph this in Desmos and get a pretty accurate answer.
Oh brings back fond memories!
x=2log3 base2
Log base 2 of 9 might be the preferred answer rather than the decimal approximation
I looked up my Chambers seven decimal place log tables and found that log 9 (to base 10) was 0.9542 (to four decimal places) and log2 was 0.3010, so 0.9542/0.3010 = approx, 3.17 for x.
Spoken like a true teacher who spends a half an hour explaining "One" question, and running off on tangents
^ --> means raise to the power of
2^x = 9 --> when searching for the exponent, logarithms are involved
Rewrite 2^x = 9 to logarithmic form log₂(9) = x
log₂(9) = x
2log₂(3) = x
-------------------------------------------------------
log₂(9) = x
log₂(3 * 3) = log₂(3) + log₂(3) = x
-------------------------------------------------------
log₂(9) = x
(log(9)) / (log(2)) = 3 ---> the base-2 log change to natural log
Just goes to show that 55 years old maths teaching still works!!!! (PS I solved it within a minute)
x.ln(2)=ln(9)= 2.ln(3)
x=2.ln(3)/ln(2)🎉
The average American has a 7th grade education. As a college physics and math tutor, I learned that many students struggle to read a textbook.
This yt uploader is probably one of them. Even if the solution is, somehow, magically correct, the delivery style does not teach an objective lesson unobstructed.
Let us assume
x=1, then 2 9
x=2, then 4 9
x=3, then 8 9
x=4, then 16 9
x will be between 3 & 4
x is not a natural number
2^3 + 1^3 = 9
You can skip ahead to 12:00 and not miss anything important.
Even someone that took algebra in high school likely never heard of natural or base 10 logs. So 90% of people would have no idea. And it wouldn't effect their lives, effect the jobs they would chose to do and want to do.
@thomassutrina7469. Logarithm was taught in my 9th grade Algebra class. I've also used it at work but I'm a retired engineer.
Question: Do you provide these in different languages?
Should have solved it using log theorems. x = log (base 2) x 9, so x = 2log (base 2) x3, so x = 3, since 2log (base 2) =1.
You can solve this just by taking a few guesses and narrowing down from there. Doesn’t take a geniuos.
2^3=8
3=log8/log2
In same way,
2^x=9
x=log9/log2
I was really hoping this video would actually talk about how to calculate a Log; that's one area of algebra that's always been a mystery to me. I get that Log 9 is a value, but HOW do you calculate it? I was hopeful until you said to take out your calculator to find the final answer. I think that would go along way to helping folks understand Logs, if we knew what the calculator was actually doing instead of it being a magic box and spitting out some never-ending decimal number. I've just never been able to visualize what its doing. Thank you.
Sorry, there is no simple calculation. Honest.
To derive the formula used to calculate log(x), you need to use calculus. The basic calculation involves summing an infinite series - in practice keep adding successive terms until you have the accuracy you require. That's why most people use a calculator or a Big book of pre-calculated values of log(x).
here is one formula for the natural log of x: ln(x) = sum (for k=1 to infinity) [ (1/k) * ((x-1)/x)^k ] valid for x >= ½
to get 15 decimal places of accuracy for ln(9) you need to evaluate about 250 terms of the form (1/k)*(8/9)^k . for k=1 to k=250 so ln(9) ~ 2.19722457733619 You need more digits, add up more terms.
@@Steven-v6l I shared that same interest in learning how to find the log; so thanks for sharing your knowledge in the reply. Your post makes a great argument for investing in a calculator!!
Ok. What? I get 2 x 2 x 2 = 8, i.e. 2 raised to the power 3. I get that! And I get 2 raised to the power 4 is 16. I get that too! But what the heck is 2 raised to the power 3.169?? 2 x 2 x 2 x 0.169? Which is, of course, wrong.
I realise that 0.169 is spread across each 2! So, dividing 0.169 by 3 (for each 2) and adding that to 2 it's very roughly something like (but isn't quite) 2.056333333 x 2.056333333 x 2.056333333 which sort of equals 9. I now understand what 2 raised to the power of 3.169 kind of is. But now you realise what you're dealing with here 😆
May I ask what application you use for the "chalkboard "?
XLog(2) = Log(9)
X = Log(9)/Log(2)
X = 3.17
(3.16992500...)
I put in 3.17. Does that count?
2^x = 9
2^x = (2)^3 + 1
2^x = (2)^3 + 2^0
xlog2 = 3log2 +0log2
x ~ 3
Starts solving at 1:30
Without a calculator, and per your first explanation wouldn't bet best answer be log2 (9) = x? I agree that using log10 (or ln) makes sense when using a calculator, but it adds a step of taking the log of both sides then doing the division, where as you can just use the identity.
Thanks your videos wake up this (almost ninety yearold brain every morning and are fun.
Man goes off on a tangent so much I get two adds between every section.
I remember solving this with logs - been a long long time
Can we get the answer without the calculator ?
That was an education
I'm adding as a new comment as well as a reply elsewhere as this is important:
Interesting.. not had to even think of logs for decades so a good revision for me. The phrase "round it off" there is very wrong. Using four figure log (tables) you can only quote the answer to three significant figures anyway but suppose you used six figure or were allowed to use a calculator then the question should either state or you should state as part of your answer the number of significant figures. So if you used a calculator say you might wish to quote to four figures giving us 3.170 and here the zero is a significant number. Extending the solution a little an astute student might point out that the rounding error is so infinitesimally small that for practical purposes the solution is 3.17
Nice 👍
Those good at math don't have any friends.
I can testify to that
3.17 = x
As a question, when I do lg9/log2 in my calculator, I get the value 3.16992500144 within the limits of my display. Wouldn't the 3 orders of magnitude precision give 3.170 instead of 3.169, as the next 9 is greater than 5, or is this a situation where because the number that one would round up from is an odd number, you don't round up? (At this level of precision, there isn't a hand tool you would use that would recognize the difference, though a machinist would probably complain.)
Why isn't this the equation: Log (base2)9=X? I ask this based on the relationship you show at the 6:05 mark. I am just following your rule for 2(exp)X=9.
Take log of both sides.
You missed out that log 9 based 10 turns in lg 9
I would use logarithms X -= log 9 / log 2
Are you using C notation? What is the initial value of X because you're specifying X is being updated by subtracting log(9)/log(2) from itself.
@@oahuhawaii2141 The negative is a mispriint.But the logarithmic equation is wrong in that it didn't specify that X=antilog(log(9)/log(2))
@@victorstalick5528: No antilog is needed. The solution is x = log(9)/log(2) = 2*log(3)/log(2) . We can approximate it using 0.47712 and 0.30103 for log(3) and log(2) respectively.
These problems are painfully fun to review. I haven't forgotten everything!
I only do the so called 4 simple maths. However I'd make a guess 4.5, now I will watch the video and see where I an wrong.. I am an old man in my 70's and I never needed this kind of math.
one could use logarithm or use "goal seek"