If \(\sqrt{\sqrt{\sqrt{3x}}} = \sqrt[4]{2x}\) , what is the greatest possible value of x?


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Key concept #1: \(\sqrt{k} = k^{\frac{1}{2}}\)

Key concept #2: \(\sqrt[4]{k} = k^{\frac{1}{4}}\)


So, take: \(\sqrt{\sqrt{\sqrt{3x}}} = \sqrt[4]{2x}\)

Rewrite as: \((((3x)^{\frac{1}{2}})^{\frac{1}{2}})^{\frac{1}{2}} = (2x)^{\frac{1}{4}}\)

Apply the power of a power law to get: \((3x)^{\frac{1}{8}}=(2x)^{\frac{1}{4}}\)

Raise both sides to the power of 8 to get: \(((3x)^{\frac{1}{8}})^8=((2x)^{\frac{1}{4}})^8\)

Apply the power of a power law to get: \((3x)^1=(2x)^2\)

Simplify to get: \(3x=4x^2\)

Rewrite as: \(4x^2-3x=0\)

Factor: \(x(4x-3)=0\)

So, EITHER \(x = 0\) OR \(4x-3=0\)

If \(4x-3=0\), then \(x=\frac{3}{4}=0.75\)

What is the greatest possible value of x?

Answer: 0.75

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2/9 ?

Nope. The question comes down to (3x)^ 1/6 = (2x) ^ 1/4

so 3x = (2x) ^6/4 or (2x)^2
3x = 4x^2
x = 3/4

it stands for (3x)^ 1/8 = (2x) ^ 1/4 ,
so 3x=(2x)^1/2,
3x=4*x^2
x=3/4

could someone explain in more detail?

\(3x^{\frac{1}{8}}= 2x^{\frac{1}{4}}\)

When both sides are raised to the fourth power

\(\sqrt{3x} = 2x\)

Squaring both the sides

\(3x = 4x^2\)

\(4x^2 - 3x = 0\)

\(x(x - \frac{3}{4}) = 0\)

\(x = 0 / \frac{3}{4}\)

Greatest possible value \(= \frac{3}{4}\)


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Thanks for the explanation!

\(\sqrt{\sqrt{\sqrt{3x}}} = ^{4}\sqrt{2x}\)

\(((3x^{\frac{1}{2}})^{\frac{1}{2}})^{\frac{1}{2}} = 2x^{\frac{1}{4}}\)

\(3x^{\frac{1}{2}*\frac{1}{2}*\frac{1}{2}} = 2x^{\frac{1}{4}}\)

\(3x^{\frac{1}{8}} = 2x^{\frac{1}{4}}\)

\((3x^{\frac{1}{8}})^{8} = (2x^{\frac{1}{4}})^{8}\)

\(3x = 4x^{2}\)

\(4x^{2} - 3x = 0\)

\(x(4x - 3) = 0\)

\(x = 0 \) and \(\frac{3}{4}\)

Greatest value of \(x = \frac{3}{4} = \)0.75

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