Essentially we are asking which is larger, 70^(1/70) or 71^(1/71). In other words, is x^(⅟x)=ˣ√x increasing or decreasing near x=70. You can use differentiation to check for slope there. Or simply note that exponentiation grows much faster than x. So for large values of x, value of ˣ√x is decreasing. So, ⁷⁰√70 > ⁷¹√71.
yeah, because the extra exponent beats the lower base. In fact the difference is about 26 times. Edit: the difference is actually exactly what he calculated: 70/e.
If you rearrange the general comparison to (x+1)/x vs Ln(x+1)/Ln(x) it seems pretty obvious that the left side will be larger since the arguments of the logs change faster than the logs. But is this always true? No, it isn't. Both expressions are equal at x = 2.293166287. Below that, the ratio of the logs is greater. Does anyone know if that number has any significance in some other context?
*@ Math Window* You made it too complicated. 71^70 vs. 70^71 Divide each side by 70^70: (71/70)^70 vs. 70^1 [(70 + 1)/70]^70 vs. 70 (1 + 1/70)^70 vs. 70 e = lim as x --> oo of (1 + 1/x)^x = 2.718... (1 + 1/70)^70 < 70 Therefore, 71^70 < 70^71.
@@Fred-yq3fs -- I just used the same step for the limit as Math Window. Do you have (an) alternative step(s) to replace what I wrote for the Olympiad Competition?
That's pretty basic stuff. A good year 11 should manage this. ln(71^70/70^71) = 70Ln(1+1/70) - ln(70) but for all x>0 we know ln(1+x)0 So: 70ln(1+1/70) < 1 As e < 70 and Ln is an increasing function, we have -ln(70)
I did it the dumb way: I compared (x+1)^x to x^(x+1). At 1 (x+1)^x is greater. At 2 also, but at 3 x^(x+1) is the greater of the two function and obviously will be greater for all greater x. So for x = 70, x^(x+1) also wins.
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Essentially we are asking which is larger, 70^(1/70) or 71^(1/71). In other words, is x^(⅟x)=ˣ√x increasing or decreasing near x=70. You can use differentiation to check for slope there. Or simply note that exponentiation grows much faster than x. So for large values of x, value of ˣ√x is decreasing.
So, ⁷⁰√70 > ⁷¹√71.
Please, How to Solve this Exercise? WHICH IS BIGGER? 17^23 or 19^13?
probably 70^71 is bigger
yeah, because the extra exponent beats the lower base. In fact the difference is about 26 times.
Edit: the difference is actually exactly what he calculated: 70/e.
Not probably brother thats a huge damn difference 💀
@@rio1628-- *Stop* with your major cursing, you rude *jerk!* It is ignorant and needless.
In fact, go back and delete that word.
@@rio1628 -- Take your major cursing *OFF* of the forum. It has no place on here.
If you rearrange the general comparison to (x+1)/x vs Ln(x+1)/Ln(x) it seems pretty obvious that
the left side will be larger since the arguments of the logs change faster than the logs. But is this
always true? No, it isn't. Both expressions are equal at x = 2.293166287. Below that, the ratio of the
logs is greater. Does anyone know if that number has any significance in some other context?
*@ Math Window* You made it too complicated.
71^70 vs. 70^71
Divide each side by 70^70:
(71/70)^70 vs. 70^1
[(70 + 1)/70]^70 vs. 70
(1 + 1/70)^70 vs. 70
e = lim as x --> oo of (1 + 1/x)^x = 2.718...
(1 + 1/70)^70 < 70
Therefore, 71^70 < 70^71.
Yeah I always start these by dividing the lesser base to the power of lesser exponent
The intermediate limit you're using is not a Y12 "known fact", so not a good proof for the competition.
@@Fred-yq3fs -- I just used the same step for the limit as Math Window. Do you have (an) alternative step(s) to replace what I wrote for the Olympiad Competition?
That's pretty basic stuff. A good year 11 should manage this.
ln(71^70/70^71) =
70Ln(1+1/70) - ln(70)
but for all x>0 we know ln(1+x)0
So:
70ln(1+1/70) < 1
As e < 70 and Ln is an increasing function, we have
-ln(70)
nice
I just compared (70)(ln(71)) and (71)(ln(70)) with a calculator.
I did it the dumb way: I compared (x+1)^x to x^(x+1). At 1 (x+1)^x is greater. At 2 also, but at 3 x^(x+1) is the greater of the two function and obviously will be greater for all greater x. So for x = 70, x^(x+1) also wins.
No, it just happened to be greater at x = 3. There is no "obviously" about it, and your statement is not a proof.