Comparing 7^6 and 6^7
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- Опубликовано: 31 июл 2024
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4th method:
Divide both by 6⁶.
You're left with (7/6)⁶ vs 6.
(7/6)⁶ = (1 + 1/6)⁶, and thus is in the form (1 + 1/n)^n, which we know is less than e (as it grows closer to e the larger the n).
Therefore:
(7/6)⁶ < e < 6
(7/6)⁶ × 6⁶ < 6 × 6⁶
7⁶ < 6⁷
look at blackpenredpen’s video on a^b vs b^a. If e
all this type if problem can be solved by reference ti the function x^(1/x) ,
a^b > b^a
take power^(1/ab)
a^(1/a)>b^(1/b)
if b>a>e
OR
divide both sides by the lower a^a. here divide by 6^6
then we compare
(7/6)^6 which is less than e
with 6 which is greater than e
thank you.
The big issue is when we have a
@ Koen Nako -- That is the [1/(ab)]th power. The denominator, ab, is inside grouping symbols.
@@robertveith6383 it doesn’t matter. Both of them mean the same unless you want to interpret it as (1/a)(b) which it’s really hard to understand why you would do that. If you really wanted that then why not just combine the power which would be (b/a)? I just don’t understand and I guess it’s more clear but WFA and most people understand what I meant. (1/ab) does not equal (1/a)(b) or (b/a).
Raise both numbers to 1/42, comparing 7^(1/7) and 6^(1/6)
Let f(x) = x^(1/x), ln f(x) = ln(x)/x
Derivative both sides:
(1/f(x)) f'(x) = (1-ln(x)) /x^2
f'(x) = (1/x^3) (1-ln(x)) ln(x)
Let f'(x)=0, ln(x) =1 or ln(x) =0, x=e or x=1
Easily see that f'(x) e, so f(x) is decreasing for all x >e > 2
So f(7) < f(6), therefore 7^6 < 6^7
I didn't even think of 'smart' ways of solving it, just went straightforwardly:
6^7 = 6 * (216)^2
7^6 = (343)^2
If we divide both the parts by (216)^2, we'll come to the following:
(343/216)^2 vs. 6
Thus, we need to compare 343/216 and the square root of 6.
But obviously, the first number is less than 2 and the second one is greater than 2, so the answer is <
Tho I agree that functioning method is more widely used and it's the way of generalization.
The first and second method are kinda identical though. I’m usually a much much bigger fan of the third one because you’re getting a more exact understanding of how big they are and you’re using the fact that the exponents are integers.
First take logarithms of both sides and compare 6 ln(7) vs 7 ln(6). Subtract and get 6(ln(7) -ln(6)) - ln (6) or 6 ln(7/6) - ln(6) or ln((7/6)^6) - ln (6).
Equivalently compare (7/6)^6 with 6. Next use the binomial theorem and calculate ((1+ (1/6))^6.
This gives 1 + ((6/((6^1)) + ((15/(6^2)) + ((20/(6^3)) + ((15/(6^4)) + ((6/((6^5)) + ((1/((6^6)) = 2 + five fractions. Note each fraction is < 1/2.
So the total is < 2 + 5 x (1/2) = 4.5 < 6. So RHS > LHS
If you consider the limit (1+1/n)^n -> e, you have 7/6 =(1+1/6), so (7/6)^6 is close to e, and much less than 6, so it follows e*6^6 < 6*6^6.
Can be solved without calculator in less than 10 seconds. 6 to power 7 is 1296*216, which is more tha 240k. 7 to power 6 is 343 * 343 which is less than 400 squared which is 160k
Whenever eb^a. Proof of that is analogical to first method. Therefore 6^7>7^6
A very interesting and beautiful point about this question.[7⁶ or 6⁷]
If we replace x with 6;
so we have: 7⁶=(x+1)^(x) and 6⁷=x^(x+1)
We divide the number 7⁶ by 6⁷
7⁶/6⁷=(x+1)^(x)/(x^(x+1))
Then we simplify it like this:
(1+/x)^(x)/(x)
OKAY, we calculate this limit:
Lim[(1+1/x)^(x)/x]
X→∞
We separate the limit:
Lim[(1+1/x)^(x)]×lim[1/x]
X→∞
Lim[(1+1/x)^(x)]=e
X→∞
Lim[1/x]=1/∞=0
X→∞
So Lim[(1+1/x)^(x)/x]=e×0=0
X→∞
Also, by plotting this function, we see that:
Lim[(1+1/x)^(x)/(x)]=1
X→2.293
So for numbers greater than 2.293
The denominator of the fraction is always larger.
So If we have a^(b) b^(a)
If 2.293
I solved using bprp's method it says if e≤ab^a
Analyse function ln(x) /x
->
Decrease When x->∞
-> 6^7>7^6
Love method 3!! 👍
I hope you all enjoy my theory I literally finished this today hahahhaa
f(x) = ln(x)/x can't be increasing on (-∞; e] because ln(x) is not defined on (-∞; 0], so does the whole function. (0; e) will do.
Wonderful
Thank you
i messed around with functions but i didn't hit on the right one
then i did it the old fashioned way
6^7 = (2^7)(3)(9^3) = (384)(9^3) > (343)(7^3) = 7^6
Great
Very easy!! Just calculate 7^6 and 6^7. However, the calculator is prohibited.
No differentiation or logarithm is required.
Then solve 999^1000 and 1000^999 which is bigger .
@@tapankoley9713 999^1000 is bigger than 1000^999.
In this case you'd better to use the differentiation of x^(1/x) and logarithm to solve it.
(for example to solve which is bigger e^π and π^e)
@@tapankoley9713 x^(x+1) is always bigger than (x+1)^x for x >2.292
if e
First method is my favorite
Great!
👍
Without prof. If x >= e, (the number of course), then x^(x+d) > (x+d)^x for any positive d always.
Russians can calculate direct answer numbers in mind faster than you try to estimate )))
Hmm 😁
Yeah
But for higher numbers
@@SyberMath the normal Russian test for 15 year scoolboy is as follows. You push-ups on hands from the floor naming next power of 2 each time. 2,4,8,16... Or you fail to up or you miss the correct number, pause is forbidden. Normal result is about 15, record I have seen 22 (4 194 304). Test is sudden, no preparation possible, loosers are punished by stick.
@@barackobama2910 Is English not your primary language? I would punish for the spelling errors.
@@robertveith6383 moreover I haven't experience for 20 years... But in 1981 I was able to remember 80 new words in 8 minutes . Teacher was strongly impressed.
学习
This question is very simple, the only knowledge it requires is to know the logarithm of numbers 1 to 10.
M: 7⁶6⁷
6Log(7)7log(6)
We know log(7)~.85 log(6)~.8
So 6(.85)=5.1 and 7(.8)=5.6
As result 7⁶
How did you know the values of log(7) and log(8).
If you use calculator, compute directly 7^6 and 6^7
@@WahranRai I said that logarithms from 1 to 10 are important for solving such problems and other problems such as calculating pH in chemistry and it is necessary to know them. So you have to learn them.
Log(1)=0
Log(2)~.3
Log(3)~.48
Log(4)=2log(2)~.6
Log(5)~.7
Log(6)=log(3)+log(2)~.78
Log(7)~.85
Log(8)=3log(2)~.9
Log(9)=2log(3)~.96~.95
Log(10)=1
Log(272)~log(270)=3log(3)+1~2.44
@@morteza3268 You must give how to approximate the values of log7 and log6.
*Thumbs-down to you, Morteza. * You should not have to know *any* logarithm approximations to do this problem! The problem could relatively easily just have asked to compare 17^15 to 13^17, etc.
@@robertveith6383 My friend, the story begins with prime numbers. But this story can also have a good ending🙂
I mean we can approximate those numbers. 17^(15) vs 13^(17)
15log(17)~15log(16)~15(4(.3))=18
17log(13)~17log(12)~17(2(.6+.48)=18.36 so 13^(17)>17^(15)
who cares about logarithms