Great problem! The approach I took was compare each log equation to establish relationships: a^6=b^4, b^4=c^3, and c^3=a^6. Then I used the change of base formula so that log(c)base(ab) = log(c)/log(ab). I then wrote that expression converting a and b so that the entire expression is in terms of c: log(c)/log(ab) = log(c)/log(c^(1/2)*c^(3/4)) = log(c)/[1/2*log(c) + 3/4*log(c)]. Then you can factor out log(c) in the denominator, log(c) then completely cancels out of the expression, and you can then directly calculate the expression: log(c)/[log(c)*(1/2 + 3/4)] = 1/(1/2 + 3/4) = 1/(5/4) = 4/5.
I took the most roundabout method you could think. turned the second eqn into 4log(ab/a) and then solved for log(c)/log(ab). after 3 pages, it came to the right answer lol.
This is called pure enjoyment.... Btw can you do like previous years maths olympiads algebra or number theory or combinatorics problems... I would love to watch those sir...
6 log a = 4 log b = 3 log c = m 6 log a = m a = 10^(m / 6 ) similarly, 4 log b = m b = 10^( m / 4 ) 3 log c = m c = 10^( m / 3 ) ab = 10^( m / 6 ) * 10^( m / 4 ) ab = 10^( m / 6 + m / 4 ) = 10^( 5m / 12 ) that means log_10^(5m / 12 ) ( 10^(m / 3 ) ) here is an interesting property of logs : log_(a^n) ( b^k ) = ( k / n ) log_a ( b ) by this property, we get : [ (m / 3 ) / ( 5m / 12 ) ] * log_10 ( 10 ) = (1 / 3 ) * ( 12 / 5 ) = 4 / 5
At first we know that; 6log[ab](a)=4log[ab](b)=3log[ab](c)=M Therefore, we add the two equalities on the left; Log[ab](b)+log[ab](a)=M(1/6+1/4) Log[ab](ab)=1=M(10/24) M=12/5 On the other hand; Log[ab](c)=M/3=4/5 THE END
My method was :- Since , 6 log a = 4 log b = 3 log c , a⁶=b⁴=c³. a⁶=c³ -> a²=c b⁴=a⁶ -> b=(a)^3/2 So,ab=(a)^5/2 log c to the base (ab)= log a² to the base ((a)^5/2), Using the logarithmic identities, log a² to the base ((a)^5/2) = 4/5 × log a to the base a = 4/5×1 = 0.8
I tried it this way: So we are given 6loga = 4logb = 3logc, If we break them up individually we have: 6loga = 3logc => a^6 = c^3 ...(1) Similarly, 4logb = 3logc => b^4 = c^3 ...(2) If we raise (2) to the power of 3/2 we get: b^6 = c^(9/2) ...(3) If we multiply (1) & (3) we get: (ab)^6 = c^(3 + 3/2) = c^(9/2) => 6log(ab) to the base (ab) = (9/2) × logc to the base (ab) => 6 = (9/2) × logc to the base (ab) => logc to the base (ab) = 12/9 = 4/3# I don't know where did I go wrong here? Please guide me. Thank you. 😇🙏
@@pratham_kun oh right! I didn't notice it... Thank you so much! I have been prone to silly mistakes since my childhood which has often cost me marks in exams.. it reminded me of the reprimands i would often receive from my teachers in my younger days.. 😂😂😂 thanks again friend! 💐😇
Setting 6 log a = 4 log b = 3 log c = 12k we have log a = 2k, log b = 3k, log c = 4k. We note that a = b = c = 1 are not suitable. Then we have k is not zero and ( log c )/( log a + log b ) = ( 4k )/( 2k + 3k ) = 4/5
Third method: As a.b=(10^x/6).(10^x/4)=10^(5x/12) Log ab=5x/12 If log c base ab= y c=ab^y that means x/3=(5x/12)^y taking the log base ab both sides Log c= log ab^y that means y= log c/ log ab Y = (x/3)/(5x/12)= 4/5 and that is the answer 😃
Before watching the video, I solved it in my mind: log c / log ab = log c / (log a + log b) 3log c = 6log a => log a = 0.5log c 3log c = 4log b => log b = 0.75log c log c / log ab = log c / 1.25 log c = 0.8
My solution went as follows: Let y = 6loga=4logb=3logc and x=log_ab(c) c=(ab)^x log(c)=x log (ab) = x(log a + log b) 6 log(c) = 2y= =x(6log a + 3/2*4log b) 2y=x(y+3/2y)=5/2xy Therefore x=5/4 If a,b,c are not all equal to 1
Great problem! The approach I took was compare each log equation to establish relationships: a^6=b^4, b^4=c^3, and c^3=a^6. Then I used the change of base formula so that log(c)base(ab) = log(c)/log(ab). I then wrote that expression converting a and b so that the entire expression is in terms of c: log(c)/log(ab) = log(c)/log(c^(1/2)*c^(3/4)) = log(c)/[1/2*log(c) + 3/4*log(c)]. Then you can factor out log(c) in the denominator, log(c) then completely cancels out of the expression, and you can then directly calculate the expression: log(c)/[log(c)*(1/2 + 3/4)] = 1/(1/2 + 3/4) = 1/(5/4) = 4/5.
After calculating that ab = c**⁵⁄4 you can use log_xⁿ(y) = 1⁄ₙ·log_x(y):
log_ab(c) = log_c**⁵⁄4(c) = ⅘·log_c(c) = ⅘.
I took the most roundabout method you could think. turned the second eqn into 4log(ab/a) and then solved for log(c)/log(ab). after 3 pages, it came to the right answer lol.
Nice! 😂
That's usually how I roll.
This is called pure enjoyment.... Btw can you do like previous years maths olympiads algebra or number theory or combinatorics problems... I would love to watch those sir...
Will get to it
6 log a = 4 log b = 3 log c = m
6 log a = m
a = 10^(m / 6 )
similarly, 4 log b = m
b = 10^( m / 4 )
3 log c = m
c = 10^( m / 3 )
ab = 10^( m / 6 ) * 10^( m / 4 )
ab = 10^( m / 6 + m / 4 ) = 10^( 5m / 12 )
that means log_10^(5m / 12 ) ( 10^(m / 3 ) )
here is an interesting property of logs : log_(a^n) ( b^k ) = ( k / n ) log_a ( b )
by this property, we get : [ (m / 3 ) / ( 5m / 12 ) ] * log_10 ( 10 )
= (1 / 3 ) * ( 12 / 5 )
= 4 / 5
Wow logs are back 🥳🥳.
This was an innovative question!
At first we know that;
6log[ab](a)=4log[ab](b)=3log[ab](c)=M
Therefore, we add the two equalities on the left;
Log[ab](b)+log[ab](a)=M(1/6+1/4)
Log[ab](ab)=1=M(10/24) M=12/5
On the other hand;
Log[ab](c)=M/3=4/5
THE END
No , in the question the given base is 10 , not ab . You are doing this by assuming that the base is ab .
@@edtx9324 Does not matter
6log(a)/log(ab)=4log(b)/log(ab)=3log(c)/log(ab)
They can be simplified:
Wow. This is pure intelligence 👏,I understood everything from beginning to the end.
My method was :-
Since , 6 log a = 4 log b = 3 log c , a⁶=b⁴=c³.
a⁶=c³ -> a²=c
b⁴=a⁶ -> b=(a)^3/2
So,ab=(a)^5/2
log c to the base (ab)= log a² to the base ((a)^5/2),
Using the logarithmic identities,
log a² to the base ((a)^5/2) = 4/5 × log a to the base a = 4/5×1 = 0.8
pfff ... log(ab)c = log c / log(ab) = log c / (log a + log b)
log a = 1/2 log c and log b = 3/4 log c
So answer is 1 / (1/2+3/4) = 4/5
😁
The second method was much smarter.
log a + log b = (3/6) log c + (3/4) log c
(4/5) log ab = log c
(ab)^(4/5) = c
therefore the sought-after quantity is 4/5
I tried it this way:
So we are given 6loga = 4logb = 3logc,
If we break them up individually we have:
6loga = 3logc => a^6 = c^3 ...(1)
Similarly, 4logb = 3logc => b^4 = c^3 ...(2)
If we raise (2) to the power of 3/2 we get:
b^6 = c^(9/2) ...(3)
If we multiply (1) & (3) we get:
(ab)^6 = c^(3 + 3/2) = c^(9/2)
=> 6log(ab) to the base (ab) = (9/2) × logc to the base (ab)
=> 6 = (9/2) × logc to the base (ab)
=> logc to the base (ab) = 12/9 = 4/3#
I don't know where did I go wrong here? Please guide me. Thank you. 😇🙏
When multiplying (1) and (3)
It should be
3^(3+9/2)=3^(15/2)......
@@pratham_kun oh right! I didn't notice it... Thank you so much! I have been prone to silly mistakes since my childhood which has often cost me marks in exams.. it reminded me of the reprimands i would often receive from my teachers in my younger days.. 😂😂😂 thanks again friend! 💐😇
@@imonkalyanbarua
🤞🤞
😂Fun fact: Silly mistakes is in my genetics. In my previous examination, i literally wrote 7-6 = -1
💀💀
@@pratham_kun 😂😂😂 keep it up! 🤪👍👍
= log c / log(ab)
= log c / (log a + log b)
Now represent denominator in log c. That's it.
Setting
6 log a = 4 log b = 3 log c = 12k
we have log a = 2k, log b = 3k, log c = 4k.
We note that a = b = c = 1 are not suitable.
Then we have k is not zero and
( log c )/( log a + log b ) = ( 4k )/( 2k + 3k ) = 4/5
= log c / (log a + log b)
= 1/(log a / log c + log b / log c)
= 1/(1/2 + 3/4)
= 1/(5/4)
= 4/5
Third method:
As a.b=(10^x/6).(10^x/4)=10^(5x/12)
Log ab=5x/12
If log c base ab= y
c=ab^y that means
x/3=(5x/12)^y taking the log base ab both sides
Log c= log ab^y that means y= log c/ log ab
Y = (x/3)/(5x/12)= 4/5 and that is the answer 😃
Before watching the video, I solved it in my mind:
log c / log ab = log c / (log a + log b)
3log c = 6log a => log a = 0.5log c
3log c = 4log b => log b = 0.75log c
log c / log ab = log c / 1.25 log c = 0.8
1st method seems shorter if you proceed logc a = log a / log c
My solution went as follows:
Let y = 6loga=4logb=3logc and x=log_ab(c)
c=(ab)^x
log(c)=x log (ab) = x(log a + log b)
6 log(c) = 2y=
=x(6log a + 3/2*4log b)
2y=x(y+3/2y)=5/2xy
Therefore x=5/4
If a,b,c are not all equal to 1
A Nice one!!! Just one remark....What if ab=1? Then log_1(c) is not defined....
Right
just think about the ratio of (log a):(log b):(log c), it is 2:3:4 obviously, and think about (ln e)/(ln a + ln b).
Bring more videos of log really very helpful 🙏
Good to hear! I love logs, too and I made quite a few log problems (my favorite pastime). Two of them are really cool and will come up in the future!
We can go with ratio and Exponential
I tried it and I came up with (ab)^(1/2) = c and thus the answer is 1/2.
i'd like to see how you came up with that answer.
Change of Base rule!!!
Why don't you ever start with the second method?
Check out the upcoming videos! 😉
Thank you
Np
very good question
Thanks, Nico! 💕
4/5
My method 2nd
thnx
Np
6/10/2022
👍👍
complicated
👍
4/5