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Solve the Following Equation: `4^(2x)=(Root3 16)^(-6/Y)=(Sqrt8)^2` - CBSE Class 9 - Mathematics

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Question

Solve the following equation:

`4^(2x)=(root3 16)^(-6/y)=(sqrt8)^2`

Solution

`4^(2x)=(root3 16)^(-6/y)=(sqrt8)^2`

`rArr4^(2x)=(sqrt8)^2` and `(root3 16)^(-6/y)=(sqrt8)^2`

`rArr4^(2x)=(8^1/2xx2)` and `(16^(1/3xx-6/y))=(8^1/2xx2)`

`rArr4^(2x)=8` and `(16^(-2/y))=8`

`rArr2^(4x)=2^3` and `(2^(-8/y))=2^3`

`rArr4x=3` and `-8/y=3`

`rArrx=3/4` and `y=-8/3`

  Is there an error in this question or solution?

APPEARS IN

 RD Sharma Solution for Mathematics for Class 9 by R D Sharma (2018-19 Session) (2018 to Current)
Chapter 2: Exponents of Real Numbers
Ex. 2.20 | Q: 16.2 | Page no. 26

Video TutorialsVIEW ALL [1]

Solution Solve the Following Equation: `4^(2x)=(Root3 16)^(-6/Y)=(Sqrt8)^2` Concept: Laws of Exponents for Real Numbers.
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