27^m+27^m+27^m m=Log[27,999]=Log[27,37]+1=Log[27,1.370 recurring]+2 Input log(27, 999) = log(27, 1 + 0.37/0.999) + 2 Result True Left hand side log(27, 999) = (3 log(3) + log(37))/log(27) Right hand side log(27, 1 + 0.37/0.999) + 2 = (3 log(3) + log(37))/log(27) It’s in my head.
Greetings. Thanks always for sharing. I did watch to the end. However, I had already determined the value of M from the expression M Log 27= 999. I am happy I watched to the end. Thanks for your hug. Blessings.
Before watching: 27 = 3^3, (a^m)^n = a^(mn). Thus you have 3^(3m) + 3^(3m) + 3^(3m) = 3(3^(3m)) = 2997. Now, why didn't I put that first 3 into the expression to get 3^(3m+1)? Because I want to divide by 3 so the RHS looks more familiar. -> 3^(3m) = 999. 999 = 333*3 = 111*9 = 37*27 = 37(3^3). Thus, 3^(3m) = 37(3^3). Taking log_3 of both sides will give us 3m = log_3(37*(3^3)) log_a(b*c) = log_a(b) + log_a(c). 3m = log_3(3^3) + log_3(37) = 3 + log_3(37) Then divide by 3... m = 1 + log_3(37)/3 If you want this in the form of a log you can plug into a TI-83+ calculator, change log_3(37) into ln (37)/ln(3) -> 1 + ln(37)/3ln(3) = 1 + ln(37)/ln(27).
27^m+27^m+27^m m=Log[27,999]=Log[27,37]+1=Log[27,1.370 recurring]+2 Input
log(27, 999) = log(27, 1 + 0.37/0.999) + 2
Result
True
Left hand side
log(27, 999) = (3 log(3) + log(37))/log(27)
Right hand side
log(27, 1 + 0.37/0.999) + 2 = (3 log(3) + log(37))/log(27) It’s in my head.
Greetings. Thanks always for sharing. I did watch to the end. However, I had already determined the value of M from the expression
M Log 27= 999. I am happy I watched to the end. Thanks for your hug. Blessings.
Correct and clear explanation. But one must review twice to master the process.
Before watching:
27 = 3^3, (a^m)^n = a^(mn).
Thus you have 3^(3m) + 3^(3m) + 3^(3m) = 3(3^(3m)) = 2997.
Now, why didn't I put that first 3 into the expression to get 3^(3m+1)? Because I want to divide by 3 so the RHS looks more familiar.
-> 3^(3m) = 999.
999 = 333*3 = 111*9 = 37*27 = 37(3^3).
Thus, 3^(3m) = 37(3^3). Taking log_3 of both sides will give us
3m = log_3(37*(3^3))
log_a(b*c) = log_a(b) + log_a(c).
3m = log_3(3^3) + log_3(37) = 3 + log_3(37)
Then divide by 3...
m = 1 + log_3(37)/3
If you want this in the form of a log you can plug into a TI-83+ calculator, change log_3(37) into ln (37)/ln(3)
-> 1 + ln(37)/3ln(3) = 1 + ln(37)/ln(27).
Thanks so much!
U are welcome
3^3+3^3+3^3=2997 1^1+^1^1+^1^3 ( m ➖ 3m+1).
There's no way that problem is from any Olympiad
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Perfect. Love your teaching beautiful and sweet lady.
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I did my dear.
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Thanks , I appreciate
I did
Wow! Thank you