x^(2log^3 x - 3/2log x)=sqrt10 (2log^3 x - 3/2log x) is the exponent solutions in my book: x1=1/6 and x2=6

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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`x^(2log^3 x - 3/2log x)=sqrt10`

Let us take log on both sides.

`(2log^3 x - 3/2log x)logx = logsqrt10`

`[2(logx)^3-3/2logx)]logx = log(10)^(1/2)`

`[2(logx)^4-3/2(logx)^2 = 1/2 log10`

 

Lets take logx = U

`2U^4-3/2U^2 = 1/2`

`2U^4-3/2U^2-1/2= 0`

`4U^4-3U^2-1 = 0`

 

`Let U^2 = t`

`4t^2-3t-1 = 0`

`4t^2-4t+t-1 = 0`

`4t(t-1)+1(t-1) = 0`

`(4t+1)(t-1) = 0`

`t = -1/4`

But `U^2 = t` so t should be `tgt=0`

t = -1/4 is not an answer.

 

then t = 1

`U^2 = 1`

`U = 1` since `U = logx`

 

So `logx = 1`

         `x = antilog1 = 10`

 

So x = 10

I think the solutions in your book are wrong. You can check this by applyong the values.

 

Sources:

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