A
B
C
D
Text Solution
Verified by Experts
The correct Answer is:
Topper's Solved these Questions
INDICES
PEARSON IIT JEE FOUNDATION|Exercise CONCEPT APPLICATION (Level 2)|20 VideosINDICES
PEARSON IIT JEE FOUNDATION|Exercise CONCEPT APPLICATION (Level 3)|7 VideosINDICES
PEARSON IIT JEE FOUNDATION|Exercise Test Your Concepts (Short Answer Type Questions)|29 VideosGEOMETRY
PEARSON IIT JEE FOUNDATION|Exercise CONCEPT APPLICATION (Level 1)|60 VideosLINEAR EQUATIONS AND INEQUATIONS
PEARSON IIT JEE FOUNDATION|Exercise CONCEPT APPLICATION (LEVEL - 3)|9 Videos
Similar Questions
Explore conceptually related problems
PEARSON IIT JEE FOUNDATION-INDICES-CONCEPT APPLICATION (Level 1)
- Find the value of (225/49)^(3//2).
Text Solution
|
- Find the value the value of 3^(2^(r^(0^(-1^(-2^(-3))))))
Text Solution
|
- 4xx(256)^((-1)/4) div (243)^(1/5)=
Text Solution
|
- Find the value of (0.000064)^(2/3) div (0.0016)^(3/4).
Text Solution
|
- If 6n=1296, then 6^(n-3) is
Text Solution
|
- The value of (9^(2)+40^(2))^(3/2)=
Text Solution
|
- [(8^(@)-7^(@))(8^(@)+7^(@))]^(0^(7^(8)))=
Text Solution
|
- Value of (125/343)^(2/3)=
Text Solution
|
- If (81)^(x)=1/((125)^(y)) and x, y are integers, then find the value o...
Text Solution
|
- If (6x)^(6)=6^(2^(3)), then find the value of x.
Text Solution
|
- Find the value of (6561)^((0.125))+(3125)^((0.2)).
Text Solution
|
- Find the value of (256)^((0.125))+(625)^((0.25))
Text Solution
|
- Evaluate the following sqrt(1/16)+(0.09)^((-1)/2)-(64)^(5/6)xx7^(@).
Text Solution
|
- If x^(y)=64, where y ne 1, then find the sum of greatest possible valu...
Text Solution
|
- Which is the greatest among (81)^(18), (243)^(15), (27)^(21) and (9)^(...
Text Solution
|
- Find the value of 4^(2^(2^(1^(3^(6))))).
Text Solution
|
- Find the value of (0.000064)^(5/6) div (0.00032)^(6/5)
Text Solution
|
- The following are the steps involved in solving the problem, if (32/24...
Text Solution
|
- Find the value of sqrt(18)+sqrt(12), if sqrt(12)=1.414 and sqrt(3)=1.7...
Text Solution
|
- (65.61)^(1/8)=.
Text Solution
|