Math, asked by shayaanlkgoxel0t, 1 year ago

3√log x + 2log √X^-1 = 2​

Answers

Answered by amitnrw
11

Given :  3√log x + 2log √x⁻¹  = 2​

To Find : Value of x

Solution:

3√log x + 2log √x⁻¹  = 2​

=> 3√log x + 2log √(1/x)   = 2​

=> 3√log x + 2log  (1/√x)   = 2​

=> 3√log x - 2log (√x)   = 2​

=> 3√log x - 2log (√x)   = 2​

=> 3√log x -  log (√x)²   = 2​

=> 3√log x -  log (x)    = 2​

=>  log (x) -  3√log x    + 2 = 0

=>  log (x) -  2√log x  -   √log x  + 2 = 0

=> √log x ( √log x - 2)  - 1 ( √log x  - 2 ) = 0

=> (√log x  - 1)( √log x  - 2 ) = 0

√log x  - 1  = 0

=> √log x = 1

=> log x  = 1  

=> x = 10  

√log x  - 2  =0

=> √log x = 2

=> log x  = 4

=> x = 10000  

x  = 10    & 10000

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