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Very nice @MATHS TUTORIAL (BL SAHU)
Thinking outside the box, 49-9=40. 7^{√m/2}=7^2, so √m/2=2 or √m=4 or m=16.Check: 7^{√16/2}=7^{4/2}=49, 3^{√16/2}=3^{4/2}=9, and 49-9=40
If m is an integer, u = (√m)/2, and x = 7^(u/2), must x be an integer? Maybe I just need caffeine, but that's the only step I can't follow.
Same question here, i don't understand that restriction. Also why in 11=7^(u/2) solution impossible? If you can not calculate it in your head, that doesn't mean that you can't calculate it at all
Note that 40=49-9 =7²+3² =[sqrt(7)]⁴-[sqrt(3)]⁴It is clear that m=16
👏👏👏and..happy good year!
Thank you.💝
😊m=4 моментально
M=16
❤❤❤❤
Very nice @MATHS TUTORIAL (BL SAHU)
Thinking outside the box, 49-9=40. 7^{√m/2}=7^2, so √m/2=2 or √m=4 or m=16.
Check: 7^{√16/2}=7^{4/2}=49, 3^{√16/2}=3^{4/2}=9, and 49-9=40
If m is an integer, u = (√m)/2, and x = 7^(u/2), must x be an integer? Maybe I just need caffeine, but that's the only step I can't follow.
Same question here, i don't understand that restriction. Also why in 11=7^(u/2) solution impossible? If you can not calculate it in your head, that doesn't mean that you can't calculate it at all
Note that 40=49-9
=7²+3²
=[sqrt(7)]⁴-[sqrt(3)]⁴
It is clear that m=16
👏👏👏and..happy good year!
Thank you.💝
😊m=4 моментально
M=16