Discussion Topic

# Solving for x in the equation (x^1/3)^(logx x^2 +2)=2log3 27

Summary:

To solve for $$x$$ in the equation $$(x^{1/3})^{(\log_x x^2 + 2)} = 2 \log_3 27$$, first simplify the right side to $$2 \cdot 3 = 6$$. Then, rewrite the left side as $$x^{(1/3)(\log_x x^2 + 2)} = x^{(2/3 + 2/3)} = x^{4/3}$$. Equate the exponents: $$\frac{4}{3} = 6$$, and solve for $$x$$ to find $$x = 729$$.

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

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

(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

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What is x if (x^1/3)^(logx x^2 +2)=2log3 27?

We have to find x if (x^1/3)^(logx x^2 +2)=2log3 27

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

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

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

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

=> (x^4/3) = 6

=> x=  6^(3/4)

The value of x is 6^(3/4)

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