Working with Logarithmic Expressions

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PLAYLISTS 🎵 :

[Skank_and_Gutterboy]

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.

[MichaelJamesActually]

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.

[Jha-s-kitchen]

Wow logs are back 🥳🥳.
This was an innovative question!

[zhugzhuangzury]

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...

[morteza]

At first we know that;

Therefore, we add the two equalities on the left;

Log[ab](ab)=1=M(10/24) M=12/5
On the other hand;
Log[ab](c)=M/3=4/5
THE END

[isaacdagwom]

Wow. This is pure intelligence 👏, I understood everything from beginning to the end.

[abhimanyusingh]

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

[michaelempeigne]

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

[jermas]

Third method:
As
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 😃

[tontonbeber]

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

[winnewFirst]

= log c / log(ab)
= log c / (log a + log b)
Now represent denominator in log c. That's it.

[danilobondi]

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

[coreyyanofsky]

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

[amoledzeppelin]

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

[rob]

= 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

[田村博志-zy]

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

[imonkalyanbarua]

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. 😇🙏

[carlosvaccaro]

1st method seems shorter if you proceed logc a = log a / log c

[monujhembrom]

Bring more videos of log really very helpful 🙏

[dmtri]

A Nice one!!! Just one remark....What if ab=1? Then log_1(c) is not defined....