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PEARSON IIT JEE FOUNDATION-INDICES-CONCEPT APPLICATION LEVEL 1
- Express 122^(2/3) as qth root of x^(p) using x^(p/q)=(x^(p))^(1/q).
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- Evaluate (16)^(5/2).
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- Evaluate a^(n).
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- (21^(2)-15^(2))^(4/3)=
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- 64^(2/3)xx64^(1/3)xx64^((-5)/3)
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- Find value of (61^(2)-11^(2))^(3/2).
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- root(6)(0.004096)=
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- [169^(-3)/((196)^(-8))]^(1/48)=
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- How do you find the value of x² + 1/x² given that x = 2+ √3?
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- 81^(1/4)xx9^(3//2)xx27^(- 4/3)=
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- If 4(4x)^(7)=4^(6^(2)), then what is the value of x ?
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- If 7^(n)=2401, then 7^(n-5)=
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- root(5)(0.03125)=
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- Find the value of (225/49)^(3//2).
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- Find the value the value of 3^(2^(r^(0^(-1^(-2^(-3))))))
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- 4xx(256)^((-1)/4) div (243)^(1/5)=
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- Find the value of (0.000064)^(2/3) div (0.0016)^(3/4).
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- If 6^n=1296, then 6^(n-3) is
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- The value of (9^(2)+40^(2))^(3/2)=
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- [(8^(@)-7^(@))(8^(@)+7^(@))]^(0^(7^(8)))=
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