Question

rewrite the following expression in simplest radical form given $x > 0$.
$sqrt3{8x} \cdot \sqrt4{256x^2}$

Explanation:

Step1: Simplify each radical separately

First, simplify \(\sqrt[3]{8x}\). We know that \(8 = 2^3\), so \(\sqrt[3]{8x}=\sqrt[3]{2^3x}=2\sqrt[3]{x}\) (since \(\sqrt[n]{a^n}=a\) for real numbers when \(n\) is odd, and here \(n = 3\) and \(x>0\)).

Next, simplify \(\sqrt[4]{256x^2}\). We know that \(256=4^4 = 2^8\), so \(\sqrt[4]{256x^2}=\sqrt[4]{2^8x^2}\). Using the property \(\sqrt[n]{a^m}=a^{\frac{m}{n}}\), we have \(2^{\frac{8}{4}}x^{\frac{2}{4}}=2^2x^{\frac{1}{2}} = 4\sqrt{x}\) (or \(4x^{\frac{1}{2}}\)). But we can also express it with exponents for easier combination later. Let's use exponents: \(\sqrt[4]{256x^2}=256^{\frac{1}{4}}x^{\frac{2}{4}}=(2^8)^{\frac{1}{4}}x^{\frac{1}{2}}=2^{2}x^{\frac{1}{2}} = 4x^{\frac{1}{2}}\). And \(\sqrt[3]{8x}=8^{\frac{1}{3}}x^{\frac{1}{3}}=(2^3)^{\frac{1}{3}}x^{\frac{1}{3}}=2x^{\frac{1}{3}}\).

Step2: Rewrite the original expression with exponents

Using the property \(\sqrt[n]{a}=a^{\frac{1}{n}}\), we can rewrite the original expression \(\sqrt[3]{8x}\cdot\sqrt[4]{256x^2}\) as \((8x)^{\frac{1}{3}}\cdot(256x^2)^{\frac{1}{4}}\).

Now, expand the exponents:
\(8 = 2^3\) and \(256=2^8\), so \((2^3x)^{\frac{1}{3}}\cdot(2^8x^2)^{\frac{1}{4}}\).

Using the property \((ab)^n=a^nb^n\), we get \(2^{3\cdot\frac{1}{3}}x^{\frac{1}{3}}\cdot2^{8\cdot\frac{1}{4}}x^{2\cdot\frac{1}{4}}\).

Simplify the exponents:
\(2^{1}x^{\frac{1}{3}}\cdot2^{2}x^{\frac{1}{2}}\).

Step3: Combine like terms (using exponent rules for multiplication)

Using the property \(a^m\cdot a^n=a^{m + n}\) for the base \(2\) and for the base \(x\) separately.

For the base \(2\): \(2^{1}\cdot2^{2}=2^{1 + 2}=2^3 = 8\).

For the base \(x\): \(x^{\frac{1}{3}}\cdot x^{\frac{1}{2}}=x^{\frac{1}{3}+\frac{1}{2}}\). To add the fractions, find a common denominator: \(\frac{1}{3}+\frac{1}{2}=\frac{2 + 3}{6}=\frac{5}{6}\), so \(x^{\frac{5}{6}}\).

So now we have \(8\cdot x^{\frac{5}{6}}\cdot\)? Wait, no, wait. Wait, when we expanded \((8x)^{\frac{1}{3}}\) we had \(2x^{\frac{1}{3}}\) and \((256x^2)^{\frac{1}{4}}\) we had \(4x^{\frac{1}{2}}\). Then multiplying \(2x^{\frac{1}{3}}\cdot4x^{\frac{1}{2}}\) gives \(8x^{\frac{1}{3}+\frac{1}{2}}\). As \(\frac{1}{3}+\frac{1}{2}=\frac{2 + 3}{6}=\frac{5}{6}\), so \(8x^{\frac{5}{6}}\). But we can also combine the radicals by finding a common index. The least common multiple of \(3\) and \(4\) is \(12\).

Let's try that approach for accuracy.

Rewrite \((8x)^{\frac{1}{3}}\) and \((256x^2)^{\frac{1}{4}}\) with denominator \(12\) in the exponents:

\((8x)^{\frac{1}{3}}=(8x)^{\frac{4}{12}}=(2^3x)^{\frac{4}{12}}=2^{\frac{12}{12}}x^{\frac{4}{12}}=2x^{\frac{1}{3}}\) (same as before)

\((256x^2)^{\frac{1}{4}}=(256x^2)^{\frac{3}{12}}=(2^8x^2)^{\frac{3}{12}}=2^{\frac{24}{12}}x^{\frac{6}{12}}=2^2x^{\frac{1}{2}}\) (same as before)

Wait, maybe better to use the property \(a^m\cdot a^n=a^{m + n}\) after expressing both with exponent form:

\(\sqrt[3]{8x}=8^{\frac{1}{3}}x^{\frac{1}{3}}=2x^{\frac{1}{3}}\)

\(\sqrt[4]{256x^2}=256^{\frac{1}{4}}x^{\frac{2}{4}}=4x^{\frac{1}{2}}\)

Now multiply them: \(2x^{\frac{1}{3}}\cdot4x^{\frac{1}{2}}=(2\times4)x^{\frac{1}{3}+\frac{1}{2}} = 8x^{\frac{2 + 3}{6}}=8x^{\frac{5}{6}}\)

But we can also express \(x^{\frac{5}{6}}\) as \(\sqrt[6]{x^5}\), but let's check the initial simplification again. Wait, maybe I made a mistake in the first approach. Let's do it by converting to 12th roots.

\(\sqrt[3]{8x}=\sqrt[12]{(8x)^4}=\sqrt[12]{8^4x^4}=\sqrt[12]{(2^3)^4x^4}=\sqrt[12]{2^{12}x^4}=2x^{\frac{4}{12}} = 2x^{\frac{1}{3}}\)

\(\sqrt[4]{256x^2}=\sqrt[12]{(256x^2)^3}=\sqrt[12]{256^3x^6}=\sqrt[12]{(2^8)^3x^6}=\sqrt[12]{2^{24}x^6}=\sqrt[12]{2^{24}}x^{\frac{6}{12}}=2^{2}x^{\frac{1}{2}} = 4x^{\frac{1}{2}}\)

Now multiply the two 12th roots:

\(\sqrt[12]{(8x)^4}\cdot\sqrt[12]{(256x^2)^3}=\sqrt[12]{(8x)^4\cdot(256x^2)^3}\)

Calculate \((8x)^4=8^4x^4 = (2^3)^4x^4=2^{12}x^4\)

\((256x^2)^3=(2^8)^3x^6=2^{24}x^6\)

Multiply them: \(2^{12}x^4\cdot2^{24}x^6=2^{12 + 24}x^{4+6}=2^{36}x^{10}\)

Now take the 12th root: \(\sqrt[12]{2^{36}x^{10}}=2^{\frac{36}{12}}x^{\frac{10}{12}}=2^3x^{\frac{5}{6}} = 8x^{\frac{5}{6}}\)

We can also write \(x^{\frac{5}{6}}\) as \(\sqrt[6]{x^5}\), but let's check if we can simplify further. Wait, maybe the problem expects a radical form with integer exponents under the radical. Wait, let's go back to the first simplification of each radical:

\(\sqrt[3]{8x}=2\sqrt[3]{x}\) and \(\sqrt[4]{256x^2}=4\sqrt[4]{x^2}\) (since \(256x^2 = 4^4x^2\), so \(\sqrt[4]{4^4x^2}=4\sqrt[4]{x^2}\))

Now multiply \(2\sqrt[3]{x}\cdot4\sqrt[4]{x^2}=8\sqrt[3]{x}\sqrt[4]{x^2}\)

To combine these radicals, we use the property \(\sqrt[n]{a}\sqrt[m]{b}=a^{\frac{1}{n}}b^{\frac{1}{m}}=a^{\frac{m}{nm}}b^{\frac{n}{nm}}=\sqrt[nm]{a^m b^n}\)

So \(\sqrt[3]{x}\sqrt[4]{x^2}=\sqrt[12]{x^4}\sqrt[12]{x^6}=\sqrt[12]{x^{4 + 6}}=\sqrt[12]{x^{10}}\)

Therefore, \(8\sqrt[12]{x^{10}}\) or \(8x^{\frac{10}{12}}=8x^{\frac{5}{6}}\)

Wait, but let's check the value of \(256x^2\) under the 4th root: \(256 = 4^4\), so \(\sqrt[4]{4^4x^2}=4\sqrt[4]{x^2}\), and \(\sqrt[3]{8x}=2\sqrt[3]{x}\). Then multiplying \(2\sqrt[3]{x}\times4\sqrt[4]{x^2}=8\sqrt[3]{x}\sqrt[4]{x^2}\). Now, \(\sqrt[3]{x}=x^{\frac{1}{3}}\) and \(\sqrt[4]{x^2}=x^{\frac{2}{4}}=x^{\frac{1}{2}}\), so multiplying the exponents: \(\frac{1}{3}+\frac{1}{2}=\frac{2 + 3}{6}=\frac{5}{6}\), so \(x^{\frac{5}{6}}=\sqrt[6]{x^5}\), so \(8\sqrt[6]{x^5}\) or \(8x^{\frac{5}{6}}\). But let's verify with actual numbers. Let \(x = 64\) (a perfect 6th power, \(64 = 2^6\)).

Original expression: \(\sqrt[3]{8\times64}\cdot\sqrt[4]{256\times64^2}\)

\(\sqrt[3]{512}\cdot\sqrt[4]{256\times4096}\)

\(\sqrt[3]{512}=8\) (since \(8^3 = 512\))

\(256\times4096=256\times4096 = 1048576\)

\(\sqrt[4]{1048576}=32\) (since \(32^4=1048576\))

Now multiply \(8\times32 = 256\)

Now using our formula \(8x^{\frac{5}{6}}\), with \(x = 64\), \(x^{\frac{5}{6}}=(64)^{\frac{5}{6}}=(2^6)^{\frac{5}{6}}=2^5 = 32\), then \(8\times32 = 256\), which matches.

If we use \(8\sqrt[6]{x^5}\), with \(x = 64\), \(\sqrt[6]{64^5}=\sqrt[6]{(2^6)^5}=\sqrt[6]{2^{30}}=2^5 = 32\), then \(8\times32 = 256\), which is correct.

Alternatively, let's do the exponent method correctly:

\(\sqrt[3]{8x}= (2^3x)^{\frac{1}{3}}=2x^{\frac{1}{3}}\)

\(\sqrt[4]{256x^2}=(2^8x^2)^{\frac{1}{4}}=2^{2}x^{\frac{1}{2}}\)

Multiply them: \(2\times2^2\times x^{\frac{1}{3}+\frac{1}{2}}=2^3\times x^{\frac{2 + 3}{6}}=8x^{\frac{5}{6}}\)

Yes, that's correct.

Answer:

\(8x^{\frac{5}{6}}\) (or \(8\sqrt[6]{x^5}\))