How do you solve $2 x^{\frac{3}{4}} = 54$ $2x^(3/4)=54$ ?

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How do you solve $2 x^{\frac{3}{4}} = 54$ $2x^(3/4)=54$ ?

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Answer 1 · 2016-07-31T08:57:30

$x = 81$ $x = 81$

Explanation:

Divide both sides by $2$ $2$ to get:

$x^{\frac{3}{4}} = 27 = 3^{3}$ $x^(3/4) = 27 = 3^3$

If $x$ $x$ is Real and positive then so is $x^{\frac{3}{4}}$ $x^(3/4)$ and we can raise both sides to the power $\frac{4}{3}$ $4/3$ to find:

$x = x^{1} = x^{\frac{3}{4} \cdot \frac{4}{3}} = {(x^{\frac{3}{4}})}^{\frac{4}{3}} = {(3^{3})}^{\frac{4}{3}} = 3^{3 \cdot \frac{4}{3}} = 3^{4} = 81$ $x = x^1 = x^(3/4*4/3) = (x^(3/4))^(4/3) = (3^3)^(4/3) = 3^(3*4/3) = 3^4 = 81$

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Other solutions?

Are there any negative or Complex solutions?

Suppose $x = r (cos θ + i sin θ)$ $x = r(cos theta + i sin theta)$ where $r > 0$ $r > 0$ and $θ \in (- π, π]$ $theta in (-pi, pi]$

Then:

$x^{\frac{3}{4}} = r^{\frac{3}{4}} (cos (\frac{3 θ}{4}) + i sin (\frac{3 θ}{4}))$ $x^(3/4) = r^(3/4)(cos ((3theta)/4) + i sin ((3 theta)/4))$

For the imaginary part to be $0$ $0$ , we must have $sin (\frac{3 θ}{4}) = 0$ $sin ((3theta)/4) = 0$

So $θ = \frac{4 π}{3} k$ $theta = (4pi)/3 k$ for some integer $k$ $k$

This can only lie in the range $(- π, π]$ $(-pi, pi]$ if $k = 0$ $k = 0$ , so $θ = 0$ $theta = 0$ .

So the only solution is the one on the positive part of the Real axis, i.e. $81$ $81$ .

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