# Find x if (x^1/3)^(logx x^2 +2)=2log3 27

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We have to find x for (x^1/3)^(log(x) x^2 +2)=2 log(3) 27

(x^1/3)^(log(x) x^2 +2) = 2 log(3) 27

=> (x^1/3)^(log(x) x^2 + 2) = 2 log(3) 3^3

=> (x^1/3)^(log(x) x^2 + 2) = 6 log(3) 3

=> (x^1/3)^(log(x) x^2 + 2) = 6

=> (x^1/3)^[(2* log(x) x + 2) = 6

=> (x^1/3)^(2+ 2) = 6

=> x^(1/3)^4 = 6

=> x^(4/3) = 6

=> (4/3) log x = log 6

=> log x = (3/4)*log 6

=> x = 10^[(3/4)*log 6]

**We get x = 10^[(3/4)*log 6]**

(x^1/3)&(logx x^2 +2) = 2log3 27

We will use logarithm and exponent properties tp solve.

First, we know that x^a^b = x^ab

==> x^(1/3)*(logx x^2 + 2) = 2log3 3^3

Now we know that log x^a = alog x

==> x^(1/3)*(2logx x + 2) = 2*3log3 3

But we know that logx x = 1 and log3 3 = 1

==> x^(1/3)*(2+2) = 6*1

==> x^4/3 = 6

Now we will raise to the power 3/4

==> **x = 6^3/4**

We'll apply the power property of exponentials:

(a^b)^c = a^(b*c)

(x^1/3)^(logx x^2 +2) = x^(logx x^2 +2)/3

But logx x^2 = 2logx x = 2

x^(logx x^2 +2)/3 = x^(2 +2)/3

x^(logx x^2 +2)/3 = x^(4/3)

x^(4/3) = 2log3 27

x^(4/3) = 2log3 (3^3)

x^(4/3) = 2*3log3 3

x^(4/3) = 6

x = 6^(3/4)

**The solution of the equation is x = 6^(3/4).**