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Good morning sir ji 🙏
Pranam guruvar 🙏🏻 🌹 ♥️
बहुत ही सुंदर ढंग से आपने सिखाया है आपने🙏🙏🙏🙏🙏🙏 lk
You gave very good information, thank you for sharing bhai 🙏🙏🙏🙏👍🏽👍🏽👍🏽 9:21 👌🏽👌🏽👌🏽👌🏽👌🏽 11:59
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🙏🙏🙏👍👍👍👏👏👏 9:42
36लाइक सर जी बहुत बढ़िया 11:55
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Radhey radhey Bhai ❤️🙏 ekadashi ki hardik shubhkamnaye bhai ❤️ sun rhi hu bhai ❤❤❤
Sorry Bhaiya hmare yaha light nhi aa rahi hai tabhi mai primiyam me nhi aa payi
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Sir. Sart me btaye time adhik hota hai
The key is • (a^b)^c=(a^c)^b=a^(bc) • 9=3², 121=11² and 33=3×11
69Ld❤
Why do you divide by × on power 3 ?
(11^×)^2=(3^×)(11^×)+(3^×)^2=>{(11/3)^×}^2=(11/3)^×)+1)=>y^2-y-1=0y={1+_ _/1+4)}/2=>y=(1+_/5)/2..&(1-_/5)/2....n.aLog@11/3 (11/3)^×=Log@11/3 (1+_/5)/2=>×=log@(11/3) (1+_/5)/2
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Good morning sir ji 🙏
Pranam guruvar 🙏🏻 🌹 ♥️
बहुत ही सुंदर ढंग से आपने सिखाया है आपने🙏🙏🙏🙏🙏🙏 lk
You gave very good information, thank you for sharing bhai 🙏🙏🙏🙏👍🏽👍🏽👍🏽 9:21 👌🏽👌🏽👌🏽👌🏽👌🏽 11:59
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Nice sharing ❤
Like done so beautiful sharing bhai Jai shree ram bahut useful video ati sundar ji 🙏🙏🙏🙏
15 like
Very nice shearing bhaiya ji 🎉🎉🎉🎉 God bless you always
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🙏🙏🙏👍👍👍👏👏👏 9:42
36लाइक सर जी बहुत बढ़िया 11:55
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Very nyc bro😊
Lk🎉
19 lk dn 😊
Radhey radhey Bhai ❤️🙏 ekadashi ki hardik shubhkamnaye bhai ❤️ sun rhi hu bhai ❤❤❤
Sorry Bhaiya hmare yaha light nhi aa rahi hai tabhi mai primiyam me nhi aa payi
25 like video running 🎉
Sir. Sart me btaye time adhik hota hai
The key is
• (a^b)^c=(a^c)^b=a^(bc)
• 9=3², 121=11² and 33=3×11
69Ld❤
Why do you divide by × on power 3 ?
(11^×)^2=(3^×)(11^×)+(3^×)^2
=>{(11/3)^×}^2
=(11/3)^×)+1)
=>y^2-y-1=0
y={1+_ _/1+4)}/2
=>y=(1+_/5)/2..&
(1-_/5)/2....n.a
Log@11/3 (11/3)^×=
Log@11/3 (1+_/5)/2
=>×=log@(11/3) (1+_/5)/2
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