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PEARSON IIT JEE FOUNDATION-INDICES -Concept Application Level-1
- 2^(3^(2))=
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- 3^(2^(0^(5)))=
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- 200000000=
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- In 2x^5, base is
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- (2^3)^4=
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- Find the value of 3^4[(2/3)^2+(2/3)-(2/(3))^3]
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- Value of ((1024)/(243))^((3)/(5)) is
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- Find the value of (2^4+2^3)^(2//6)
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- (25^2-15^2)^((3)/(2))=
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- (33^2-31^2)^((5)/(7))=
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- If abc =0 then find the value of [(x^a)^b]^c.
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- If a+b+c=0, then find the value of sqrt(x^a.x^b.x^c).
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- Simplify (sqrt(36)+sqrt(64))/(2^3-1).
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- If a=25 ,then find the value of a^(25^(0))+a^(0^(25)).
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- Find the value of (-1)^(301)+(-1)^(302)+(-1)^(303)+.......+(-1)^(400).
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- If 3^x=6561, then 3^(x-3) is
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- The following steps are involved in finding the value (7+x)^3 , when (...
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- The following steps are involved in finding the value of ((x^a)/(x^b))...
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- The following steps are involved in finding the value of 3^(n-3) , whe...
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- The following steps are involved in finding the value of (a^(x+y))^(x-...
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