Comparing 5^23 and 7^15

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[adandap]

5^4 = 625 > 343 = 7^3 then raise both sides to the 5th power. 5^20 > 7^15

[davidseed]

Another method.
I know that the function x^(1/x) has a max at x=e.
So for integers>2, if y>x then
y^(1/y) < x^(1/x)
Raising both tp the power of xy x^y > y^x
(The power wins)
So 5^7>7^5
Cubing
5^21>7^15
Thus 5^23>7^15

[yoganandasp]

Bro compare 5^23 vs 7^19 not 16 plz these are very close

[jmadratz]

Write
7^z = 5
Take the log of each side
And rearrange
Z= log5/log7=0.827…
Substitute for 5=7^0.827…
7^0.827*23 or 7^15
7^19.0… or 7^15
Obviously
Left side greater (5^23)

[ardademirhan]

10>7 and 5^23>10^15=(5^15)*2^15 since 5^8> 4^8=2^16>2^15.

Many other methods are possible, 23 and 15 have nice successors: 24 and 16, 24=1.5*16 but 5^(1.5) >7 since sqrt 125 > sqrt 49 so 5^24 =5^(1.5*16)> 7^16. Now to get 5^23 and 7^15 we divide LHS by 5 and RHS by 7 so we are making the RHS even smaller (even RHS/5 would be less than LHS/5, so RHS/7 is of course smaller than LHS/5). Done.

Lets see why 5^24 > 7^18. 5^4>7^3 hence for their sixth powers 5^24>7^18, divide both by 5 we get 5^23> (0.2)*7^18>7^17>7^16 so we got a stronger result than your open ended question.

The fact that 5^4 / 7^3 is almost 2 allows us stronger results. Lets compare 5^24 and 7^20. 5^6 vs 7^5 will be equivalent comparison. 15625 vs 16807, the 7 guy wins but only by a small margin. So 5^23<7^20. Lets see why (16807/15625)^4 < 1.4=7/5 hence 7^20>5^23>7^19. 16807<16900 and 15625>15600 so it suffices to show that (16900/15600)^4 =(13/12)^4 is less than 7/5. The last part is ugly but no calculus or calculators needed! (169^2)*5 = 5*10, 000+69*200*5+5*69^2=119, 000+5*69^2; (144^2)*7= 7*10, 000+44*200*7+7*44^2=131, 600+7*44^2. 5*(69^2-44^2)=5*(25*113)=5*2, 825=14, 125. As 131, 600-119, 000=12, 600, it suffices to show that 2*44^2 >14, 125-12, 600=1525 but 40^2=1600 so it is clear (computing 44^2 was very easy also but not needed). Hence we have seen that 5^23 beats 7^19 but loses to its father 7^20.

[alexeyrb]

Nice solution in the video. I went with a different one. 5*2 = 10, 7*sqrt(2) is also very close to 10 (certainly between 9.8 and 10.2), because sqrt(2) is 1.4 with a bit (14*2 = 196, very close to 200).
Given that, multiply both sides by 2^23. On the left side, we end up with 10^23, on the right we have 7^15 * 2^23 = 7^15 * 2^(7.5) * 2^(15.5) ~ 10^15 * 2^15.5.
Now, dividing both sides by 10^15, we are comparing 10^8 to 2^15.5 = 32768*sqrt(2) < 50000. The left-hand side being at least 2000 times large, even though we've made a slight rounding off with 7*sqrt(2), there is no way that difference, even raised to the 15th power covers the 2000 multiplicative factor. The gap is so large that it's won't be too difficult to come up with an easy-to-compute upper bound on (10/sqrt(2)*7)^15.
Quick, but kinda sloppy.

[euler]

I worked out the first few powers of 5 and 7 ; and determined that 5^5 > 7^4

Raising both sides to the power 4 gives

5^20 > 7^16 ; again you can construct a series of inequalities

5^23 > 5^20 > 7^16 > 7^15

[jonathantoler]

I did 7^2 = 50 vs 5^2= 25 as the basic simplification. Remove and replace as you go

Do that seven times and you get the seven side being 2^7 times 7 versus 5^8 then simply to the 4 if needed but it should be obvious at this point the 5^23 was bigger to begin with

[Skank_and_Gutterboy]

Before watching the vid or doing any analysis, I'm guessing that 5^23 is larger. The base shouldn't matter much and 23 is a significantly higher power than 15.

[kurtlichtenstein]

5^18 <7^15 <5^19
As for 7^16, it's easy to prove it's less than 5^20, and fun to prove it's more than 5^19.

[RisetotheEquation]

Syber, if you had a gun pointed at my head and said I have 1 minute to decide (BANG if I'm wrong), there's NO WAY I would figure this out in time playing with the bases and the exponents (great trick, BTW!). Here's what I'd do to try to stay alive in time without a calculator:
23log5 vs 15log7
23/15 vs log7/log5
≈1.5 vs log5/log5 = 1
≈1.5 vs log25/log5 = 2
No way log7/log5 is going to close the gap in time to get you above 1.5
So I go with 5^23.
(And say a little prayer that I'm right!)

[raivogrunbaum]

5^23>4^23=2^46>2^45=8^15>7^15

[Hjerpower]

i used 50>49 for my solution but yours is a lot easier to work out

[jlloberaq]

7²~50 = 2*5² but 2¹⁰~10³ => 2⁷~5³.
With this information, 7¹⁵ = 7*(7²)⁷ ~ 7*2⁷*5¹⁴ ~ 7*5¹⁷ << 5²³.
Even though there is no inequality, this factor 5⁴-5⁵ leads us unequivocally to 7¹⁵<5²³.

[allanmarder]

Take the 5th root of both numbers and compare (5^23)^(1/5) with (7^15 )^(1/5) i.e. 5^4.6 with 7^3. Now 5^4 equals 625 and 7^3 equals 343 so LHS > RHS

[daviebowie]

There are so many ways to take this comparison.

Mine is 5^3>7^2 and 5^2>7.

[MizardXYT]

Calculating the first few terms of the continued fraction for ln(7)/ln(5), you get [1; 4, 1, 3] = 1+1/(4+1/(1+1/3)) = 23/19. This means ln(7)/ln(5) ≈ 23/19, or 7¹⁹ ≈ 5²³, which implies 7¹⁵ < 5²³.

If you want to compare 7¹⁹ and 5²³, you can look at the sequence of longer and longer continued fraction approximations. The approximation will cycle between greater and less than the target value.
[1] = 1 < ln(7)/ln(5)
[1;4] = 5/4 > ln(7)/ln(5)
[1;4, 1] = 6/5 < ln(7)/ln(5)
[1;4, 1, 3] = 23/19 > ln(7)/ln(5)
23 ln(5) > 19 ln(7)
5²³ > 7¹⁹

[Fcruibees]

Note that 5*23=7*15=105
We can look at the other side of this question: For any N is natural number, we separate it as x+x+……+x in n terms, then let f(x)=x^n, find x if f(x) has max.
The conclusion is when x=e f(x) has max. Also, if |x-e| either increase or decrease, f(x) is decreasing
Back to the question, since 5 is more approach to e, so 5^23>7^15

[tychophotiou]

Much easier method is to divide both by 5^15.
Then we are comparing 5^8 with 1.4^15.... A sinch!!! 😉

[ІгорСапунов]

we can rewrite this comparing: 23/15 ? log5(7). 7^5>5^6, then 6<log5(7)<7. Multiplying to 3: log5(7)<21<23