Comparing log_10(11) and log_11(12)
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My solution is like this.
10->11: percentage increase is 1/10
11->12: percentage increase is 1/11
1/10is greater than 1/11, so log(10)11is greater.
Base e: ln11/ln10 ~ ln12/ln11 so we compare ln²11 and ln10∙ln12. But by AM-GM: √(ln10∙ln12) < (ln10+ln12)/2 = ln√120 < ln√121 = ln11. Hence log₁₀11 > log₁₁12
EDIT: this can be generalized if we set 11=x: √ln(x+1)ln(x-1) < (ln(x+1)+ln(x-1))/2 = ln√(x²-1) < ln√x² = ln(x), assuming that x>1
x must be greater than 2 because ln (x-1) must be positive . And also, we must observe the positive value of inside the root.
Just the proof I was looking for, thanks.
Honestly, this is more a logic question than a maths question.
You are suppose to REASON why the correct answer is what it is, not calculate it mathematically.
Good thinking!
I started following this sir from today
🤩
Thank you for your alternative solution,sir I solved tricky way which is based on A.M>=G.M. If I explain,
121>120 => log 121>log120=> log11>(log10+log12)/2 > (log 10*log12)^(0.5)
=> (log11)^2>log 10*log12=> (log11/log10)>(log12/log11)
A Must-see: th-cam.com/video/5dKy5IQZy0Y/w-d-xo.html
Tare! Problema complexa si cu schepsis! Bravo! Next?!
Logarithms are very important in Analytical Chemistry that is why I teach my students on them. Usualy, I draw graphs of two of them -- one with a base of 10, the other of e.
For x > 1, that with a higher base is always the lower graph (easily to proof mathematically). That is why the point (11, log10(11)) is on the higher graph and that (12, log11(12)) on the lower.
We can see the derivative of the function:
f=log (x+1) with base x.
We can see it will come decreasing function for all x>e. So log 11, base 10 will be greater.
It decreasing for any x>0 with V.A at x=1 for example log2(3)>log2.5(3.5)
@@yoav613 Ohh yes, you are right mate...
You had calculated mentally the derivative and deduced its sign ?!
@@WahranRai Yess, with very less paper work too.
Your method is very short , Sharma Ji .Keep it up !God bless you .
You are genius , dear professor .Keep it up !
Thank you, I will
I don't understand anything but I feel like gaining intelligence +.5
🤩
Let log₁₀11 =a, log₁₁12=b, then 11=10^a, 12=11^b so 11+1=11^b , 10^a+1=11^b from here (10+1)^a- F(a)=11^b, 11^a-F(a)=11^b, 11^a>11^b, a>b. Maybe I missed something, but think this is right and short way of solution....
y = log x = ln x /ln10
y' = 1/( x ln10 ) > 0 for all x > 0
y'' = -1/( xx ln10 ) < 0 for all x > 0
Therefore, y is increasing but the rate of increase is decreasing as x increases,
For the three numbers 10, 11, 12,
log11 - log10 > log12 - log11 > 0
0 < log10 < log11
We get,
( log11 - log10 )/log10 > ( log12 - log11 )/ log11
log11/log10 - 1 > log12/log11 - 1
This is a good problem from SyberMath Sir 😃
The comparison simply follows from the concavity of the log function.
Nice!
Hi, late to the party but I found an algebraic method
Put A=log(11), B=log11(12). (log is to base 10)
By base conversion formula
A-B = log(11) - log(12)/log(11)
log(11)*(A-B) = log(11)^2 - log(12)
Use log(10) = 1
log(11)*(A-B) = log(11)^2 - log(10)*log(12)
= log(11)^2 - log(11/(11/10)) * log(11*(12/11))
= log(11)^2 - [log(11) - log(11/10)] * [log(11) + log(12/11))
Simplify
log(11)*(A-B) = log(11) * [log(11/10) - log(12/11)] + log(11/10)*log(12/11)
= log(11)*log(121/120) + log(11/10)*log(12/11)
> 0
There is more numbers betwen 11 and 121 than 10 and 100 hence, more I would guess that log base 10 of 11 is bigger than log base 11 of 12
Both log(11) and log_11(12) are greater than 1. Let's look at log_11(12). Change this formula as 1/log_12(11). It interchanges the base and the argument. Taking the reciprocal of that number makes the value smaller. Therefore, log(11)>log_11(12).
When i solve kind of comparings i switched logs to fraction
Log10(11) 》11\10 and
Log11(12) 》 12/11
Now when we compare them we can see that 11/10 is greater so log10(11) also is greater
What does that even mean sorry..the fraction of two logs does.not generally equal thenlog of a fraction remember..so I'm assuming that's not what you meant
@@leif1075so sorry my eng is so bad that's why you don't understand me i mean that if logs are assigned as fractions (not equal) then their comparing will be the same
Because we can write log10(11) as ln10/ln11 so ln10 is smaller than ln11 then 10 will be also smaller than 11
@@dionysus1210 -- No, log [base 10] (11) = [ln(11)]/[ln(10)] from the change of base rule. It is the
log of the argument divided by the log of the former base.
A Must-see: th-cam.com/video/5dKy5IQZy0Y/w-d-xo.html
nice
Thanks
Good evening sir
Good Evening
11/10>12/11
you confuse me when you say log 12 with base 11
What should I say instead?
@@SyberMath idk but how about we just use ln's so that there is no worries
@@michaelempeigne3519 /woooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooosh
@@SyberMath at 3:59 do you realize all that g(×) is unnecessary ? Since inntje numerator younhwve x tikes ln x but as x increases that means ln x
@@SyberMath log 11 of 12
log 11 is greater