I used the 4th method, which is brute forcing and substituting x with log base 12 of 18. Methods like that are universal. They generally work, but sometimes things get a little complicated to deal with.
log_12 18 = log_12 1,5 + 1. since 12 = 8*1.5 with the change of base formula, we got 1/(log_1,5 8 + 1) 😁 obviously, this can be writen as 1/(ln 8/ln 1,5 + 1) = 1/[ln 8/(ln3 - ln2) + 1] =[ (ln3 - ln2)/ 3ln2 + 1]^-1 = [(ln3 + 2ln2)^-1 but it's more fun to substitute x = 1/(log_1,5 8 + 1) you'll get (2x - 1)/(x - 2) = (2 - log_1,5 8 - 1) over (1 - 2log_1,5 8 + 2) = (1 - log_1,5 8) over (3 - 2log_1,5 8) = (log_1,5 1,5/8) over [log_1,5 (1,5/4)^3]. now, time to use change of base formula. But let use base 1,5/4 🤣 : (log_1,5 1,5/8) over [log_1,5 (1,5/4)^3] = [(log 1,5/4) + (log 1/2)] over 3 = (1 + log 1 - log 2)/3 = (1 - log 2)/3 so simple right ? but not so simple because the log is in base 1,5/4 😅 now, use change of base formula again. (1 - log 2)/3 = {1 - (1/log_2 1,5/4)}/3 = {1 - (1/[log_2 1,5 - 2])}/3 oops, I probably miss something 😭😭
Before watching the video: Given: 12↑x = 18 To find: 2↑((2x - 1)/(x - 2)) Isolating prime factors on both sides to isolate 2: (2↑2x)·3↑x = (3↑2)·2 Isolating all factors of 2: (2↑2x)/2 = (3↑2)/3↑x using (a↑m)/(a↑n) = a↑(m - n) to combine exponents: 2↑(2x - 1) = 3↑(2 - x) Inverting the base of RHS: 2↑(2x - 1) = ⅓↑(x - 2) Exponentiating both sides by (1/(x - 2)) to obtain required LHS: 2↑((2x - 1)/(x - 2)) = ⅓
I didn't use logarithms at all. I took the equation 12^x = 18 and multiplied both sides by 12^(-2) to get 12^(x - 2) = 1/8. Then I raised both sides to the 1/(x - 2) power to get 12 = 1/8^(1/(x - 2)), from which we get 8^(1/(x - 2)) = 1/12. Now, we know that 2^((2x - 1)/(x - 2)) is the same as 2^(2 + 3/(x - 2)), which equals 4 times 2^(3/(x - 2)), also known as 4 times 8^(1/(x - 2))... using the earlier result for the quantity 8^(1/(x - 2)) of 1/12, the answer is 4 times this, which is 4/12, which reduces to 1/3. No logs.
@@SyberMath I did something else I took X= Log 18 to base 12 Then substituted values Then I took - 1 as - log 12 to base 12 Then used property of log Similarly in denominator also Then I got something log to base 2 Then again identity Got the answer
I did it in a childish way, actually I first thought the problem was gonna be a challenging one. Have to find 2^{(2x - 1) /( x-2)} I did, 2^{(2x - 4)/(x-2) + 3/(x-2)} That means I have to find, 4 × 2^{3/(x-2)} 12^x =18, 12^(x-2)=18/144 12= 2^{- 3/(x-2)} 2^{3/(x-2)}=1/12 Therefore, the answer is, 4×1/12 =1/3
The first thing I did was do a prime factorization of 12^x = 3^x * 2^2x = 18 2^(2x-1/x-2) = (2^2x-1)^1/x-2. Focusing on 2^2x-1 = 2^2x/2. But 2^2x = 18/3^x. By substitution, 2^2x/2 = 18/(2*3^x) = 9/3^x = 3^2/3^x = 3^(2-x). So now we have 2^(2x-1/x-2) = (2^2x-1)^1/x-2 = 3^(2-x/x-2) = 3^-(x-2/x-2) = 3^-1 = 1/3.
*@ Robert L. Underwood* -- Your post is *thoroughly wrong.* You are missing required grouping symbols in several places. Your post needs to be heavily edited.
12^x=18 (3)^x(2²)^x=(3²)(2) So here, why can't we compare the powers? as both 3 and 2 are prime numbers. We get: x=2 if we compare powers of 3 in LHS and RHS. We get: x=½ if we compare powers of 2 in LHS and RHS. My question is, why am i getting wrong value of x? Where did I go wrong? Please help.
Very confusing method 1 I'll see what #2 method WOW dude 🙄 you need to SLOW down and explain better!! I'm new to your tutorial, so I guess I need to get use to your speed but really you need to SLOW your roll!!
I used the 4th method, which is brute forcing and substituting x with log base 12 of 18. Methods like that are universal. They generally work, but sometimes things get a little complicated to deal with.
log_12 18 = log_12 1,5 + 1. since 12 = 8*1.5 with the change of base formula, we got 1/(log_1,5 8 + 1) 😁
obviously, this can be writen as 1/(ln 8/ln 1,5 + 1) = 1/[ln 8/(ln3 - ln2) + 1] =[ (ln3 - ln2)/ 3ln2 + 1]^-1 = [(ln3 + 2ln2)^-1 but it's more fun to substitute x = 1/(log_1,5 8 + 1)
you'll get (2x - 1)/(x - 2) = (2 - log_1,5 8 - 1) over (1 - 2log_1,5 8 + 2) = (1 - log_1,5 8) over (3 - 2log_1,5 8) = (log_1,5 1,5/8) over [log_1,5 (1,5/4)^3]. now, time to use change of base formula. But let use base 1,5/4 🤣 :
(log_1,5 1,5/8) over [log_1,5 (1,5/4)^3] = [(log 1,5/4) + (log 1/2)] over 3 = (1 + log 1 - log 2)/3 = (1 - log 2)/3 so simple right ? but not so simple because the log is in base 1,5/4 😅 now, use change of base formula again. (1 - log 2)/3 = {1 - (1/log_2 1,5/4)}/3 = {1 - (1/[log_2 1,5 - 2])}/3
oops, I probably miss something 😭😭
I did it without evaluating x. I've tried using laws of exponents and number factorization and the 2nd method gets me right!!! (Ans.: 1/3) Easy...
Before watching the video:
Given:
12↑x = 18
To find: 2↑((2x - 1)/(x - 2))
Isolating prime factors on both sides to isolate 2:
(2↑2x)·3↑x = (3↑2)·2
Isolating all factors of 2:
(2↑2x)/2 = (3↑2)/3↑x
using (a↑m)/(a↑n) = a↑(m - n) to combine exponents:
2↑(2x - 1) = 3↑(2 - x)
Inverting the base of RHS:
2↑(2x - 1) = ⅓↑(x - 2)
Exponentiating both sides by (1/(x - 2)) to obtain required LHS:
2↑((2x - 1)/(x - 2)) = ⅓
I just wanna say this is beautiful
that's i found too. Have a nice day.
it feels like i learn new things everyday
Glad to hear that! 🥰
That's a good thing
That's to replace the many things that get deleted everyday. LOL
I always enjoy you presenting different approaches, awesome.
Glad to hear it!
Syber, just beautiful. Wow! You're an excellent mathematician! 👏🏻
Aww, thank you for the kind words!!! 💕
@@SyberMath well, of course. Not only are you a talented mathematician, but an excellent teacher
@@titan1235813 I appreciate that! 🥰
I liked method 2 the most, it was simple and elegant, although my mind went automatically to a combination between methods 1 & 3.
Did it by last method, uninspiring brute force but second method way more elegant wish I had discovered it. Love how you present different approaches
Glad to hear that 🙂
Nice watching this 💪
I didn't use logarithms at all. I took the equation 12^x = 18 and multiplied both sides by 12^(-2) to get 12^(x - 2) = 1/8. Then I raised both sides to the 1/(x - 2) power to get 12 = 1/8^(1/(x - 2)), from which we get 8^(1/(x - 2)) = 1/12. Now, we know that 2^((2x - 1)/(x - 2)) is the same as 2^(2 + 3/(x - 2)), which equals 4 times 2^(3/(x - 2)), also known as 4 times 8^(1/(x - 2))... using the earlier result for the quantity 8^(1/(x - 2)) of 1/12, the answer is 4 times this, which is 4/12, which reduces to 1/3. No logs.
Before watching video, my answer is 1/3
Let's check answer
Edit : Yayyy I am correct
Nice!
@@SyberMath I did something else
I took X= Log 18 to base 12
Then substituted values
Then I took - 1 as - log 12 to base 12
Then used property of log
Similarly in denominator also
Then I got something log to base 2
Then again identity
Got the answer
Thanks!
using a base-2 log and setting y = log-base-2-of 3 makes it pretty straightforward.
Love your 2nd method!!
Thank you!
I did it in a childish way, actually I first thought the problem was gonna be a challenging one.
Have to find 2^{(2x - 1) /( x-2)}
I did, 2^{(2x - 4)/(x-2) + 3/(x-2)}
That means I have to find, 4 × 2^{3/(x-2)}
12^x =18, 12^(x-2)=18/144
12= 2^{- 3/(x-2)}
2^{3/(x-2)}=1/12
Therefore, the answer is, 4×1/12 =1/3
nice methods sybermath
I enjoy watching all these smart ideas solving unusual problems. best Mathematics channel to watch.
Glad you like them, Samir!
And thank you for the kind words!!! 🥰
The first thing I did was do a prime factorization of 12^x = 3^x * 2^2x = 18
2^(2x-1/x-2) = (2^2x-1)^1/x-2. Focusing on 2^2x-1 = 2^2x/2. But 2^2x = 18/3^x. By substitution, 2^2x/2 = 18/(2*3^x) = 9/3^x = 3^2/3^x = 3^(2-x).
So now we have 2^(2x-1/x-2) = (2^2x-1)^1/x-2 = 3^(2-x/x-2) = 3^-(x-2/x-2) = 3^-1 = 1/3.
*@ Robert L. Underwood* -- Your post is *thoroughly wrong.* You are missing required grouping
symbols in several places. Your post needs to be heavily edited.
12^x=18
(3)^x(2²)^x=(3²)(2)
So here, why can't we compare the powers? as both 3 and 2 are prime numbers.
We get: x=2 if we compare powers of 3 in LHS and RHS.
We get: x=½ if we compare powers of 2 in LHS and RHS.
My question is, why am i getting wrong value of x? Where did I go wrong? Please help.
This only works when the exponents are rational numbers
@@anshumanagrawal346 Good point!
very nice question
Thanks
I got the second solution :)
Əla həll etdiniz.Bakıdan salamlar.Azərbaycan.
Salamlar! 🥰
2nd method is too good
how do you do this ? on an ipad ?
Yes
@@SyberMath and what are your grades ?
I solved it using the 2nd method. 😊
Great 👍
@@SyberMath 🤗
👌👍
I was able to work it out in my head that the answer was 1/3
hii sir
Hii!
All I can say is 👏👏👏👏
🥰 Thank you!
1/3 ans
👏👏👏👏👏
👍
1/3
1/3 wil the answer
Excellent as usual.
I appreciate that!
1/3 ...... in 5 minutes
Very confusing method 1
I'll see what #2 method
WOW dude 🙄 you need to SLOW down and explain better!!
I'm new to your tutorial, so I guess I need to get use to your speed but really you need to SLOW your roll!!
Root 6
faurmidable.think you!!!
Np. Thanks!
1/3