The main idea was to use the laws of logarithms and then some multiplication and division. It was a good problem to wreslte with for any 12th-grade student.
I actually did work through the system of equations with substitution, and I got to the right answer in the end, but it was pretty painful. I didn't see the trick to get a value for xyz, which was clearly the key to simplifying everything. I also converted to the squared bases, just to avoid having to deal with the fractions during conversion. The values come out the same.
With proper factoring, you get the linear equation Ma=b where M is the matrix [[1 1 2][1 2 1][2 1 1]] (all ones except the off diagonal), 'a' is the vector [log(x) log(y) log(z)], and b is the vector [log(81) log(16) log(1)]. Computing the inverse matrix and solving for 'a' gives you that a=[log(1/6) log(8/3) log(27/2)], thus the arguments of the logs in 'a' are the values for x,y, and z.
all of the equations can be expressed in the following forms by raising to the highest base: xy(z^2) = 81 x(y^2)z = 16 (x^2)yz = 1 we have xyz = 6, hence, x = 1/6 y = 8/3 z = 81/6
Step 1: Make the base the same for each equation: log9 x + log9 y + log9 z² =2 log16 x + log16 y² + log16 z = 1 log25 x² + log25 y + log25 z = 0 Step 2: Combine logarithms using log x + log y = log xy: log9 xyz² = 2 log16 xy²z = 1 log25 x²yz = 0 Step 3: Remove logarithms using logb x = y => x = b^y: xyz² = 9^2 = 81..................(1) xy²z = 16^1 = 16................(2) x²yz = 25^0 = 1...................(3) Step 4: solve for y using the equation (3): y = 1/(x²z)..............(4) Step 5: Substitute y = 1/(x²z) into equation (1) and solve for z: z = 81x................(5) Step 6: Substitute this to equation (4): y = 1/(81x³).............(6) Step 7: Substitute y = 1/(81x³) and z = 81x to equation (2): x * 1/(81x³)² * 81x = 16 => x^4 = 1/(81*16) => x = 1/(3*2) = 1/6. Solution x = -1/6 won't do. All x, y and z must be positive because otherwise the logarithms aren't defined. Step 8: Calculate y and z using (5) and (6): x = 1/6 y = 8/3 z = 27/2 Checking these with the second equations indicates correct values.
I tried converting all logs to the common one, and it gave me: lnx + lny + 2lnz = 4*ln3; lnx + 2lny + lnz = 4*ln2; 2lnx + lny + lnz = 0; this system isn't so tedious to solve. But the presented solution in the video is more ellegant, I must say.
I think the "trick" is to recognize that multiplying them all together leads to a nice equation where both sides are 4th powers, and then you can quickly make progress. I spent a lot of time trying to add and subtract and factor before watching the video, and it just took so long to get anywhere. But if you multiply everything it simplifies nicely.
Yo this is so cool! My friend did the Euclid contest and he gave me his sheet of problems and I was actually able to solve this particular problem, small world!
Neat solution. Wolfram Alpha concurs. I solved the three simultaneous equations in the traditional manner, by first eliminating z and then eliminating y.
by the systems 1)xyz²= 81 , 2)xy²z=16 and 3)x²yz=1. Dividing 1) for 2), 2) for 3) and 3) for 1), respectively, we have the relations z/y=81/16, y/x=16 and x/z=1/81. -> z= 81x and y =16x. Using one of the systems: xyz² =x*16x*(81x)²=2^4*9^4*x^4=3^4 -> 2*9*x=3 -> x=1/6, y = 8/3, z=27/2. q.e.d.
We can have 9 to the power of the first equation, 16 to the power of the second equation, and 25 to the power of the third equation. It’s an alternative solution and may be simpler to understand.
I really like this problem and your video! 🤩 I used a different approach at the beginning which I find a little easier. In each equation I took the largest number of the log bases (e.g. 9 in the first equation). Then I raised this number to the equation itself. E.g. the first line will then become: 9^[log₉x + log₉y + log₃z] = 9^2 Using the exponent rule that sums in the exponents represent a product of the individual exponents, we get: 9^log₉x * 9^log₉y * 9^log₃z = 81 The first two factors simplify neatly to x and y respectively, but in order to simplify the third factor, I rewrote 9 as 3²: x * y * (3²)^log₃z = 81 In a case like this, we can "swap" the exponents and turn (3²)^log₃z into (3^log₃z)². Now, with matching bases this factor simplifies to z², and we get the same equation you found: xyz² = 81 😀 I really like the "trick" to multiply all three equations and divide each line by it. I didn't see that but took the old-fashioned approach to eliminate one variable after the other, which actually isn't all that complicated here. I started with the third equation because the product is 1 and solved for z: x²yz = 1 z = 1/(x²y) Then plugged this one into the second and solved for y: xy²z = 16 (xy²) / (x²y) = 16 y/x = 16 y = 16x And with these two you can use the first equation to find the x. (I'll spare you the calculation.) I don't quite understand where your hang-up was because there's no difficulty here but I'm happy to that you found this much more elegant way, multiplying all equations. The only criticism would be that you ignored the other solutions. Just like you got x⁴y⁴z⁴ = 6⁴, my method yielded that x⁴ = (1/6)⁴. In both cases we get a positive and a negative solution, as well as two complex ones (not sure if the problem was looking only for the reals). And just as you should then have divided each line with 6, -6, 6i, and -6i, I found the values for y and z using the four solutions for x, i.e. 1/6, -1/6, i/6, and -i/6. Again, I really enjoyed this video. I hope the channel will keep growing as you deserve more viewers!
Correct. (xyz)^4 = 6^4 (xyz)^4 - 6^4 = 0 ((xyz)^2 - 6^2) * ((xyz)^2 + 6^2) = 0 Second factor has no real solutions First factor can be further simplified to (xyz - 6)(xyz + 6) = 0 so xyz = ±6
In hindsight no. If xyz were negative it would mean that either one or all three were negative while log functions don't use negative numbers (if I'm not wrong).
1:01 ok i am back lets continu learning a bit 1:05 nice i like the intro! 1:56 but all the other numbers in each equantion are the squarroot of the other both that is surely not just coincidence 3:16 ok knoweing that give you a chance to solve it. 3:38 yes if i would hat learnt that it would be fine :D 4:32 i also had done this step! 5:01 wait why can we do this? 6:06 ok just repiting? 6:44 ok we have 3 variables and 3 equantions that should be solveable. 8:49 that sounds funny lets write it down 9:15 ok i know this way of solveing equention but that wouldn´t come to my mind in this situation 9:44 looks like we should take the 4. root of it on both sides 9:59 no but you can rewrite it as (3*2*1)^4 = (xyz)^4 and than we knowe 6^4 = (xyz)^4 10:14 that is that what i thought but don´t we need a plusminus here ? 11:03 ok that is quite easy now 12:04 it realy would take to long 12:33 but why? 14:19 never say never but ok xD 17:28 hopefully! LG K.Furry
0:34 nice why of thinking! 0:56 no i have never worked with the log funktion! i have to comment the rest later because i have to make a short/long break
Namaste sir, I am one of ur subscriber from India and you content is awesome and very helpful for us. I have solved this same question earlier
Me too from india ❤
I actually did work through the system of equations with substitution, and I got to the right answer in the end, but it was pretty painful. I didn't see the trick to get a value for xyz, which was clearly the key to simplifying everything. I also converted to the squared bases, just to avoid having to deal with the fractions during conversion. The values come out the same.
With proper factoring, you get the linear equation Ma=b where M is the matrix [[1 1 2][1 2 1][2 1 1]] (all ones except the off diagonal), 'a' is the vector [log(x) log(y) log(z)], and b is the vector [log(81) log(16) log(1)]. Computing the inverse matrix and solving for 'a' gives you that a=[log(1/6) log(8/3) log(27/2)], thus the arguments of the logs in 'a' are the values for x,y, and z.
I find watching you proceed to solve such problems is always comforting!
all of the equations can be expressed in the following forms by raising to the highest base:
xy(z^2) = 81
x(y^2)z = 16
(x^2)yz = 1
we have xyz = 6, hence,
x = 1/6
y = 8/3
z = 81/6
bruh simplify
Step 1: Make the base the same for each equation:
log9 x + log9 y + log9 z² =2
log16 x + log16 y² + log16 z = 1
log25 x² + log25 y + log25 z = 0
Step 2: Combine logarithms using log x + log y = log xy:
log9 xyz² = 2
log16 xy²z = 1
log25 x²yz = 0
Step 3: Remove logarithms using logb x = y => x = b^y:
xyz² = 9^2 = 81..................(1)
xy²z = 16^1 = 16................(2)
x²yz = 25^0 = 1...................(3)
Step 4: solve for y using the equation (3):
y = 1/(x²z)..............(4)
Step 5: Substitute y = 1/(x²z) into equation (1) and solve for z:
z = 81x................(5)
Step 6: Substitute this to equation (4):
y = 1/(81x³).............(6)
Step 7: Substitute y = 1/(81x³) and z = 81x to equation (2):
x * 1/(81x³)² * 81x = 16 => x^4 = 1/(81*16) => x = 1/(3*2) = 1/6.
Solution x = -1/6 won't do. All x, y and z must be positive because otherwise the logarithms aren't defined.
Step 8: Calculate y and z using (5) and (6):
x = 1/6
y = 8/3
z = 27/2
Checking these with the second equations indicates correct values.
I tried converting all logs to the common one, and it gave me: lnx + lny + 2lnz = 4*ln3; lnx + 2lny + lnz = 4*ln2; 2lnx + lny + lnz = 0; this system isn't so tedious to solve. But the presented solution in the video is more ellegant, I must say.
I think the "trick" is to recognize that multiplying them all together leads to a nice equation where both sides are 4th powers, and then you can quickly make progress. I spent a lot of time trying to add and subtract and factor before watching the video, and it just took so long to get anywhere. But if you multiply everything it simplifies nicely.
Yo this is so cool! My friend did the Euclid contest and he gave me his sheet of problems and I was actually able to solve this particular problem, small world!
This one was easy but fun. That's the magic of Mathematics.
Neat solution. Wolfram Alpha concurs. I solved the three simultaneous equations in the traditional manner, by first eliminating z and then eliminating y.
Nice
by the systems 1)xyz²= 81 , 2)xy²z=16 and 3)x²yz=1. Dividing 1) for 2), 2) for 3) and 3) for 1), respectively, we have the relations z/y=81/16, y/x=16 and x/z=1/81. -> z= 81x and y =16x. Using one of the systems: xyz² =x*16x*(81x)²=2^4*9^4*x^4=3^4 -> 2*9*x=3 -> x=1/6, y = 8/3, z=27/2. q.e.d.
Thanks 👍
We can have 9 to the power of the first equation, 16 to the power of the second equation, and 25 to the power of the third equation. It’s an alternative solution and may be simpler to understand.
I love your channel, do you like physics.?
Just the easy part
exactly the way i did it !
I really like this problem and your video! 🤩
I used a different approach at the beginning which I find a little easier.
In each equation I took the largest number of the log bases (e.g. 9 in the first equation). Then I raised this number to the equation itself. E.g. the first line will then become:
9^[log₉x + log₉y + log₃z] = 9^2
Using the exponent rule that sums in the exponents represent a product of the individual exponents, we get:
9^log₉x * 9^log₉y * 9^log₃z = 81
The first two factors simplify neatly to x and y respectively, but in order to simplify the third factor, I rewrote 9 as 3²:
x * y * (3²)^log₃z = 81
In a case like this, we can "swap" the exponents and turn (3²)^log₃z into (3^log₃z)². Now, with matching bases this factor simplifies to z², and we get the same equation you found:
xyz² = 81 😀
I really like the "trick" to multiply all three equations and divide each line by it. I didn't see that but took the old-fashioned approach to eliminate one variable after the other, which actually isn't all that complicated here.
I started with the third equation because the product is 1 and solved for z:
x²yz = 1
z = 1/(x²y)
Then plugged this one into the second and solved for y:
xy²z = 16
(xy²) / (x²y) = 16
y/x = 16
y = 16x
And with these two you can use the first equation to find the x. (I'll spare you the calculation.)
I don't quite understand where your hang-up was because there's no difficulty here but I'm happy to that you found this much more elegant way, multiplying all equations.
The only criticism would be that you ignored the other solutions.
Just like you got x⁴y⁴z⁴ = 6⁴, my method yielded that x⁴ = (1/6)⁴. In both cases we get a positive and a negative solution, as well as two complex ones (not sure if the problem was looking only for the reals).
And just as you should then have divided each line with 6, -6, 6i, and -6i, I found the values for y and z using the four solutions for x, i.e. 1/6, -1/6, i/6, and -i/6.
Again, I really enjoyed this video. I hope the channel will keep growing as you deserve more viewers!
dovrebbe esserci un'altra soluzione reale con xyz = -6
Nice nice problem :) Great presentation.
x = 1/6
y = 8/3
z = 27/2
Oh,you are a great mathematician .Keep it up !!!
I think you missed 3 sets of solutions. Even staying with reals, xyz could also be -6.
Correct.
(xyz)^4 = 6^4
(xyz)^4 - 6^4 = 0
((xyz)^2 - 6^2) * ((xyz)^2 + 6^2) = 0
Second factor has no real solutions
First factor can be further simplified to
(xyz - 6)(xyz + 6) = 0
so xyz = ±6
In hindsight no. If xyz were negative it would mean that either one or all three were negative while log functions don't use negative numbers (if I'm not wrong).
@@MadaraUchihaSecondRikudo yes, but all the original logs are defined only for positive arguments, x,y,z > 0, then you can't have x.y.z = -6.
Excellent description on logarithms.
Love your methods bro, keep it up 🙌
Excellent Sir.
yoo i actualy joined this contest this year, i remember this question lol, this is one of the 5/10 questions that i can do
Sachin sir from pw also give this que in class
I remember this question from when i took this test lol 😅
😊👍
Fantastic video! Very clean solution
Thank you kind sir!
😊 thanks I solved it in a complicated way but got the same answer 😅
1:01 ok i am back lets continu learning a bit
1:05 nice i like the intro!
1:56 but all the other numbers in each equantion are the squarroot of the other both that is surely not just coincidence
3:16 ok knoweing that give you a chance to solve it.
3:38 yes if i would hat learnt that it would be fine :D
4:32 i also had done this step!
5:01 wait why can we do this?
6:06 ok just repiting?
6:44 ok we have 3 variables and 3 equantions that should be solveable.
8:49 that sounds funny lets write it down
9:15 ok i know this way of solveing equention but that wouldn´t come to my mind in this situation
9:44 looks like we should take the 4. root of it on both sides
9:59 no but you can rewrite it as (3*2*1)^4 = (xyz)^4 and than we knowe 6^4 = (xyz)^4
10:14 that is that what i thought but don´t we need a plusminus here ?
11:03 ok that is quite easy now
12:04 it realy would take to long
12:33 but why?
14:19 never say never but ok xD
17:28 hopefully!
LG
K.Furry
0:34 nice why of thinking!
0:56 no i have never worked with the log funktion!
i have to comment the rest later because i have to make a short/long break