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m^7 = (2^7)^8 = (2^8)^7. So, m=2^8 = 256
1. Let √(m√(m√(m)))=m^(1/2)*m^(1/4)*m^(1/8)=m^(7/8)=128=2^72. Both sides to the power of 8, m^7=2^(7*8)=256^73. Thus m=256
Тоже так решил задачу
From inside m^(1+1/2)=m^(3/2) Now sqrt m^(3/2)=m^(3/4).Now sqrt(m*m^3/4)=sqrt(m^(7/4)=m^(7/8)Therfore m^(7/8)=128=2^7=> m=(2^7)^(8/7)=2^8.
Complicated approach. Power 7 is same bases are equal. Hence answer is 2 power8
Very lnteresting Observe minutely
Koi short trik btao
sqrt( m * sqrt( m * sqrt( m ))) = 128sqrt( sqrt( m^3 * sqrt( m ))) = 128sqrt( sqrt( sqrt( m^7 ))) = 128m^(7/8) = 128m^(7/8) = 2^7m = 2^(7 * 8/7) = 2^8 = 256
Excelent
Not able to write 8 properly
And if there are a hundred roots of m, can you square this equation a hundred times? Is there enough paper?
m = 2^8 = 256
Ans is 2^8.
Thanks 👍
nice method
No great feat of math-talent here!
Why doing it in a round about way? In the end, could have used a^m^n identity as well.
Lengthy process
Ese número 8 es horrible😂😂😂😂😂😂
Good. Thanks so much
m^7= 16384^4...m=256
😮
Muito bom
m=256
m=2
Not correct method Very lengthy
I have never seen such writing method of '8'. Completely wrong style indeed. You may correct yourself.
Ok thanks for comment
I write like this, my so many friends write like this
Harvard?
16.
Id est 2× 4^1/5 respondeo. 😅😊😂😂
Id est 2× 4^1/5 respondeo. 😅😊😂😅😊🎉🎉🎉😅😊
Is this 10th standard problem or 8th standard NTSE level problem and now that's what Harvard is? Comeon this problem stinks
Worst 8 ever
m =64/7
m(¹/²+¹/⁴+¹/⁸)=m⁷/⁸=2⁷m=2⁸
thanks for comment
Simply bogus type solution .
M=2
m^7 = (2^7)^8 = (2^8)^7. So, m=2^8 = 256
1. Let √(m√(m√(m)))=m^(1/2)*m^(1/4)*m^(1/8)=m^(7/8)=128=2^7
2. Both sides to the power of 8, m^7=2^(7*8)=256^7
3. Thus m=256
Тоже так решил задачу
From inside m^(1+1/2)=m^(3/2)
Now sqrt m^(3/2)=m^(3/4).
Now sqrt(m*m^3/4)=sqrt(m^(7/4)=m^(7/8)
Therfore m^(7/8)=128=2^7
=> m=(2^7)^(8/7)=2^8.
Complicated approach. Power 7 is same bases are equal. Hence answer is 2 power8
Very lnteresting
Observe minutely
Koi short trik btao
sqrt( m * sqrt( m * sqrt( m ))) = 128
sqrt( sqrt( m^3 * sqrt( m ))) = 128
sqrt( sqrt( sqrt( m^7 ))) = 128
m^(7/8) = 128
m^(7/8) = 2^7
m = 2^(7 * 8/7) = 2^8 = 256
Excelent
Not able to write 8 properly
And if there are a hundred roots of m, can you square this equation a hundred times?
Is there enough paper?
m = 2^8 = 256
Ans is 2^8.
Thanks 👍
nice method
No great feat of math-talent here!
Why doing it in a round about way? In the end, could have used a^m^n identity as well.
Lengthy process
Ese número 8 es horrible
😂😂😂😂😂😂
Good. Thanks so much
m^7= 16384^4...m=256
😮
Muito bom
m=256
m=2
Not correct method
Very lengthy
I have never seen such writing method of '8'. Completely wrong style indeed. You may correct yourself.
Ok thanks for comment
I write like this, my so many friends write like this
Harvard?
16.
Id est 2× 4^1/5 respondeo. 😅😊😂😂
Id est 2× 4^1/5 respondeo. 😅😊😂😅😊🎉🎉🎉😅😊
Is this 10th standard problem or 8th standard NTSE level problem and now that's what Harvard is? Comeon this problem stinks
Worst 8 ever
m =64/7
m(¹/²+¹/⁴+¹/⁸)=m⁷/⁸=2⁷
m=2⁸
thanks for comment
Simply bogus type solution .
M=2