Glad you enjoyed it! Thank you! Cheers! So nice of you, Prasoon You are awesome. Keep it up 😀
@أبوأنساليمني-م8ح
2 жыл бұрын
شرح تعليمي ممتاز جداً، نتابعك من الشرق الأوسط من أقدم حضارة تاريخية من اليمن.
@wtspman
2 жыл бұрын
I solved it using log base 10. First I converted 0.5^b = 2^(-b). 1) log (5^a) = log 1000; alog5 = 3; a = 3/(log5); 1/a = (log5)/3 2) log (2^(-b)) = log 1000; -blog2 = 3; -b = 3/(log2); -1/b = (log2)/3; 3) 1/a - 1/b = (log5 + log2)/3; 1/a - 1/b = log(5*2)/3; 1/a - 1/b = (log10)/3; 1/a - 1/b = 1/3
@PreMath
2 жыл бұрын
Fabulous Thank you! Cheers! You are awesome. Keep it up 😀
@RashmisABCD
2 жыл бұрын
Fantastic share, beautifully done and shown this video, have a wonderful day, lk8
@PreMath
2 жыл бұрын
Many thanks, Rashmi Excellent! Cheers! So nice of you You are awesome. Keep it up 😀
@namshup9689
2 жыл бұрын
I really got amazed by myself today. I never thought that i could sole this problem, and it's hard quite hard for me. However, I can solve it in 30 minutes thinking. Thank you professor for beautiful and good exercises!
@PreMath
2 жыл бұрын
Excellent, Nam You are most welcome! Thank you! Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA!
@stupidhuman897
2 жыл бұрын
the presentation can be simpler, if we write 0.5 as 1/10 x 5 and take log at the very beginning. log 1000 =3 a = 3/log5 b = 3/log(1/10 x5) = 3/ (log5 -1) 1/a and 1/b have the common denominator 3 and log5 just cancelled out. the whole stuff can be worked out in mind.
@anatoliy3323
2 жыл бұрын
Both methods are wonderful. The first method is simpler whereas the second one is more interesting. I prefer the second one. Thank you so much, Professor! Nice to watch your video lessons again and again. All the best to you and to your family!
@PreMath
2 жыл бұрын
You are most welcome! Thank you! Cheers! So nice of you, Anatoliy You are amazing and very kind. Keep it up 😀 Love and prayers from the USA! Stay safe!
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