Solving a Nice Homemade Exponential Equation

[@paulschumacher1263]

Is this the first time you have started with the Second Method? I haven't watched all your videos, but many of them, and I have never seen you do that before. And if you present it first, is it really the second?

[@77Chester77]

No, it was not the first time. Usually the first is more brute-force and the second one uses a trick

[@notlin1976]

Today I prefer the 2nd method... lol Seriously, both are very interesting. 🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷🇧🇷

[@DownDance]

I enjoy these math exercises for fun so much!

[@kanankazimzada2500]

Perfect solution. You are the best in this channel

[@JimLambier]

I appreciate the pace of your videos. In particular, you don't dwell on every tiny arithmetic operation.

[@bbt584]

First, I collected a and b on one side, then after simplifications, I found 6^b=3^a. Using logarithm, a/b=logarithm 3^a/b=6 is output instead of 3^a/b=6

[@nikolakosanovic9931]

First means that it is done before every other, second means that it is done after first So in witch ever order you wrote it it will be first then second then third and so on

[@georget8008]

I followed a different path, by using the prime number theorem by I ended up in a dead end. I did the following: I facrorized the two sides of the initial equation into product of powers of 2 and 3 as you have done. the 2 and 3 are primes, so according to the PNT their exponents are unique. thus, I equate the exponents of the 2s and the exponents of the 3s. I end up with a simple, linear system of 2×2 equations. However, if tou solve this system you end up with a=b=0. But b cannot be zero, because in such case a/b couldn't be defined. where is the mistake? can anybody help?

[@musicsubicandcebu1774]

LHS base is 12 and RHS is 18 . . . so their powers cannot be equated.

[@georget8008]

​​@@musicsubicandcebu1774 the pnt says that each integer can be factored in a unique set of prime factors raised to a power. if you take the equation at 1:42 and Instead of cross dividing, you set the exponent of 2 on the left side, equal to the exponent of 2 on the right and then do the same with the exponents of 3, then you get the 2×2 system I am talking about.

[@musicsubicandcebu1774]

@@georget8008 I think you mean the 'fundamental theorem of arithmetic" not 'prime number theorem' . . . fta means that 72, as an example, can be written as 2³3² . Here, the powers of 2 are different and cannot be equated. I hope I'm understanding you correctly.

[@0ZkYtEKE]

Neither 12^(a+b) nor 18^(2a-b) are integers, so the prime factorization theorem does not apply. I made the same mistake at first.

[@georget8008]

I found out what was my mistake. By using the prime number theorem, I made the implicit assumption that the two sides of the initial equation are integers. So, I have restricted the values of a and b to numbers that make them integers.

[@mathboy8188]

Nice problem, but there's one small detail missing. 12^( a + b ) = 18^( 2a - b ) implies 6^b = 3^a is true enough, but from there you can't say that 3^(a/b) = 6, because it's possible that b = 0. That's the case when a = b = 0, which does satisfy the original statement 12^( a + b ) = 18^( 2a - b ). So you need to include the requirement that b not equal to 0 to make this problem 100% valid.

[@lukamedin6986]

I'm not clear which log property does equal log in the final step of the first method 6?

[@Ulissescars]

You can simplify because 3 is to the power of a log of base 3 and it becomes only the 6

[@scottleung9587]

Nice solution, although I thought in the beginning you said it wouldn't be so nice!

[@qwang3118]

Since a/b appears in the question, the implicit assumption is b 0. Collect a and b to different sides, yield 3^(3a) = 6^(3b). Taking 1/(3b) power, 3^(a/b) = 6. Too trivial, no interesting technical insight. ...

[@prasantbhattarai9337]

Bro‚ What's mean by the home made eqtn?

[@tambuwalmathsclass]

The equation is created by him.

[@SyberMath]

That means if someone steals it, I”ll know! 😂

[@tambuwalmathsclass]

@@SyberMath 🤣🤣🤣🤣

[@maxpetrochenko5025]

NFT equation

[@timewalkwalker]

Such an amazing channel I found question like these great

[@adalchami]

Nice!!! thanks

[@praveensomashekar82]

Can a and b both be 0?

[@hhhh82user]

No. The first step would be true, but 3^(0/0) is undefined because (0/0) is undefined.

[@sybermaths]

You are the best my teacher

[@MrLidless]

Your aversion to natural logs is disturbing, though in this case you can use log or ln. Slightly smoother 2nd method: (a + b)ln12 = (2a - b)ln18 (a + b)(2ln2 + ln3) = (2a - b)(2ln3 + ln2) 3aln3 = 3b(ln3 + ln2) (a / b)ln3 = ln3 + ln2 = ln6 3^(a / b) = 6

[@oniondesu9633]

in number theory it is very common to write log even when referring to the natural log

[@SyberMath]

No aversion. Same thing

[@mateusandradedeoliveira4034]

bro, your videos are so fucking nice. thansk for make them

[@RayyRayy2013]

6^3b=3^3a

[@tetsuyaikeda4319]

suddenly lnX is appearing in my sight

[@HoSza1]

x/y=k, (1-x)/(1-y)=? Express the latter value in terms of k!

[@ghost-jp2in]

Done without using log within 1 min😁😁😄😄

[@Ki8kiki88]

6

[@lawsofmaths429]

Thanks for sharing such an informative vedio.i also made vedio with easy tricks.

[@giuseppemalaguti435]

6

[@mathswan1607]

6