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]
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(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).
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