In cases like this where the base and exponents differ by one I’ve seen the number with the larger exponent be the bigger term every time. I expect there to be a case where this does not apply, but I don’t know what it is. If you think about it tho, 10^x is always less than 9^(x+1) for all positive integers. In fact, the difference diverges rather quickly. Whether that is true for y^x and (y-1)^(x+1) for all positive values of X and y… I dunno.
My solution: let's divide by 9^10, then we have to compare (10/9)^10 vs. 9, or (1+1/9)^10 vs.9. Let us take a square root (thanks to @stolaire !), then we have to compare (1+1/9)^5 vs. 3. (1+1/9)^5 = 1^5+5*(1^4)*(1/9)+10*(1^3)*(1/9^2)+10*(1^2)*(1/9^3)+5*1*(1/9^4)+1/(9^5) = 1+5/9+10/81+10/(9^3)+5/(9^4)+1/(9^5) = 1+(9*5+10)/81+(81*10+9*5+1)/(9^5) = 1+55/81+856/(9^5). Each of the two last summands is
General Approach: To compare a^b and c^d:
1. Formulate the Ratio:
(a^b) / (c^d)
If (a^b) / (c^d) > 1
then a^b > c^d
If (a^b) / (c^d) < 1
then a^b < c^d
# 2. Reformulate:
(a^b) / (c^d) = (a/c)^b * (1 /c^(d-b))
# Example:
1. Substitute Values:
(10^10) / (9^11)
2. Reformulate:
(10^10) /(9^11) = (10/9)^10 *(1/9)
# 3. Simplify:
(1+1/9)^10*(1/9) < 1.
# 4. Conclusion:
Since (10^10) / (9^11) < 1, it follows that 10^10 < 9^11
Thus, 9^11 is indeed greater than 10^10.
10^10
😮
In cases like this where the base and exponents differ by one I’ve seen the number with the larger exponent be the bigger term every time. I expect there to be a case where this does not apply, but I don’t know what it is. If you think about it tho, 10^x is always less than 9^(x+1) for all positive integers. In fact, the difference diverges rather quickly.
Whether that is true for y^x and (y-1)^(x+1) for all positive values of X and y… I dunno.
My solution: let's divide by 9^10, then we have to compare (10/9)^10 vs. 9, or (1+1/9)^10 vs.9. Let us take a square root (thanks to @stolaire !), then we have to compare (1+1/9)^5 vs. 3.
(1+1/9)^5 = 1^5+5*(1^4)*(1/9)+10*(1^3)*(1/9^2)+10*(1^2)*(1/9^3)+5*1*(1/9^4)+1/(9^5) = 1+5/9+10/81+10/(9^3)+5/(9^4)+1/(9^5) = 1+(9*5+10)/81+(81*10+9*5+1)/(9^5) = 1+55/81+856/(9^5). Each of the two last summands is
I just imagined the graph of ln x and thought ln 10 is bigger than ln 9, bbut it is definitely not 11/10 times bigger
I did it in my head.
10^10 : 9^11
10^10
=(9×10/9)^10
=(9^10)(10/9)^10
9^11
=(9^10)×9
10^10 : 9^11
=(10/9)^10 : 9
(10/9)^5
How about this?
10^10 ??? 9^11
sqrt(10^10) ??? sqrt(9^11)
10^5 ??? 9^(5 + 0.5)
100000 ??? 9*9*9*9*9*sqrt(9)
100000 ??? 81*81*27
100000 ??? 6561*27
10000 ??? 6561*2.7
6666*1.5 + 1 < 6561*2.7
Rough, but solves the problem, but not universal
You have noticed that sqrt(9) is integer, it's cool! so we can decrease the power by taking the sqrt, then it's easier to compare