Power Line of 7 (with Questions) - Chapter 2 Class 8 Ganita Prakash - Power Lines

part 2 - Power Line of 7 - Power Lines - Chapter 2 Class 8 - Power Play (Ganita Prakash) - Class 8 (Ganita Prakash & Old NCERT)
part 3 - Power Line of 7 - Power Lines - Chapter 2 Class 8 - Power Play (Ganita Prakash) - Class 8 (Ganita Prakash & Old NCERT) part 4 - Power Line of 7 - Power Lines - Chapter 2 Class 8 - Power Play (Ganita Prakash) - Class 8 (Ganita Prakash & Old NCERT) part 5 - Power Line of 7 - Power Lines - Chapter 2 Class 8 - Power Play (Ganita Prakash) - Class 8 (Ganita Prakash & Old NCERT) part 6 - Power Line of 7 - Power Lines - Chapter 2 Class 8 - Power Play (Ganita Prakash) - Class 8 (Ganita Prakash & Old NCERT) part 7 - Power Line of 7 - Power Lines - Chapter 2 Class 8 - Power Play (Ganita Prakash) - Class 8 (Ganita Prakash & Old NCERT)

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Power Line of 7Power line of 7 looks like Power is decreasing as we go down β– (7^7&-&823543@7^6&-&117649@7^5&-&16807@7^4&-&2401@7^3&-&343@7^2&-&49@7^1&-&7@7^0&-&1@7^(-1)&-&1/7@7^(-2)&-&1/49@7^(-3)&-&1/343@7^(-4)&-&1/2401)We need to answer some questions 2,401 Γ— 49? 493 = ? 343 Γ— 2,401 = ? (πŸπŸ”,πŸ–πŸŽπŸ•)/πŸ’πŸ— = ? πŸ•/πŸ‘πŸ’πŸ‘ = ? (πŸπŸ”,πŸ–πŸŽπŸ•)/(πŸ–,πŸπŸ‘,πŸ“πŸ’πŸ‘) = ? 1,17,649 Γ— 𝟏/πŸ‘πŸ’πŸ‘ = ? 𝟏/πŸ‘πŸ’πŸ‘ Γ— 𝟏/πŸ‘πŸ’πŸ‘ = ? Let’s answer them one by one 2,401 Γ— 49? 2,401 Γ— 49 = πŸ•^πŸ’ Γ— πŸ•^𝟐 = πŸ•^(πŸ’+𝟐) = 7^6 = 1,17,649 493 = ? 493 = γ€–(πŸ•^𝟐)γ€—^πŸ‘ = πŸ•^(𝟐 Γ— πŸ‘) = 7^6 = 1,17,649 πŸ•/πŸ‘πŸ’πŸ‘ " = ?" 7/343 = 7^1/7^3 = πŸ•^(πŸβˆ’πŸ‘) = 7^(βˆ’2) = 𝟏/πŸ’πŸ— (πŸπŸ”,πŸ–πŸŽπŸ•)/(πŸ–,πŸπŸ‘,πŸ“πŸ’πŸ‘) = ? 16,807/8,23,543 = 7^5/7^7 = πŸ•^(πŸ“βˆ’πŸ•) = 7^(βˆ’2) = 𝟏/πŸ’πŸ— 343 Γ— 2,401 = ? 343 Γ— 2,401 = πŸ•^πŸ‘ Γ— πŸ•^πŸ’ = πŸ•^(πŸ‘+πŸ’) = 7^7 = 8,23,543 (πŸπŸ”,πŸ–πŸŽπŸ•)/πŸ’πŸ— = ? 16,807/49 = 7^5/7^2 = πŸ•^(πŸ“βˆ’πŸ) = 7^3 = 343 1,17,649 Γ— 𝟏/πŸ‘πŸ’πŸ‘ = ? 1,17,649 Γ— 1/343 = πŸ•^πŸ• Γ—πŸ•^(βˆ’πŸ‘) = 7^(7 + (βˆ’3)) = πŸ•^(πŸ• βˆ’ πŸ‘) = 7^4 = 2,401 𝟏/πŸ‘πŸ’πŸ‘ Γ— 𝟏/πŸ‘πŸ’πŸ‘ = ? 1/343 " Γ— " 1/343 = πŸ•^(βˆ’πŸ‘) Γ— πŸ•^(βˆ’πŸ‘) = 7^(βˆ’3 + (βˆ’3)) = πŸ•^(βˆ’πŸ‘ βˆ’ πŸ‘) = 7^(βˆ’(3+ 3)) = πŸ•^(βˆ’πŸ”) = 1/7^6 = 𝟏/(𝟏,πŸπŸ•,πŸ”πŸ’πŸ—)

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 15 years. He also provides Accounts Tax GST Training in Delhi, Kerala and online.