Radical versus rational exponents

[d_schaeffer]

I have had the question below posed to me in class and I cannot seem to figure out why rational exponents have advantages over radical symbols (nor can I come up with any examples because I don't know how it relates). I did read in our book that they're easier to solve with a calculator, but I'm not sure that's what she's looking for. Does anyone have any suggestions?

While the radical symbol is widely used, converting to rational exponents has advantages. Explain an advantage of rational exponents over the radical sign. Include in your answer an example of an equation easier to solve as a rational exponent rather than a radical sign.

Would GREATLY appreciate any help/suggestions anyone can give.

 

[Denis]

Example where easier to solve:

n^p * n^p = n^4
n^(p+p) = n^4
n^(2p) = n^4
2p = 4
p = 2

 

[soroban]

Hello, d_schaeffer!

Why do rational exponents have advantages over radical symbols?

Example: \(\displaystyle \L\,\sqrt{x}\cdot\sqrt[3]{x}\)

Only "like" radicals can be combined.

So we modify them like this: \(\displaystyle \,\begin{array}{cc}\sqrt{x}\,=\,\sqrt[6]{x^3} \\ \sqrt[3]{x}\,=\,\sqrt[6]{x^2}\end{array}\)

And the problem becomes: \(\displaystyle \L\,\sqrt[6]{x^3}\cdot\sqrt[6]{x^2} \:=\:\sqrt[6]{x^3\cdot x^2} \;= \;\sqrt[6]{x^5}\)


With rational exponents, we have: \(\displaystyle \L\,x^{\frac{1}{2}}\cdot x^{\frac{1}{3}}\;=\;x^{\frac{1}{2}+\frac{1}{3}}\;=\;x^{\frac{5}{6}} \;= \;\sqrt[6]{x^5}\)

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Example: \(\displaystyle \L\:\frac{x}{\sqrt[5]{x^3}}\)

Rationalize: \(\displaystyle \L\:\frac{x}{\sqrt[5]{x^3}}\cdot\frac{\sqrt[5]{x^2}}{\sqrt[5]{x^2}}\;=\;\frac{x\cdot\sqrt[5]{x^2}}{\sqrt[5]{x^3\cdot x^2}} \;= \;\frac{x\cdot\sqrt[5]{x^2}}{\sqrt[5]{x^5}} \;= \;\frac{x\cdot\sqrt[5]{x^2}}{x} \;= \;\sqrt[5]{x^2}\)


With rational exponents: \(\displaystyle \L\:\frac{x}{x^{\frac{3}{5}}}\;=\;x^{1-\frac{3}{5}}\;=\;x^{\frac{2}{5}} \;= \;\sqrt[5]{x^2}\)

 

[pka]

Here is a history lesson for you.
The nth root of number a, \(\displaystyle \sqrt[n]{a}\) is itself a number, the nth power of which is a.
That is \(\displaystyle \left( {\sqrt[n]{a}} \right)^n = a\). This being the case, what notation should we use to express \(\displaystyle \sqrt[n]{a}\) in exponential form?

Well we want to preserve the regular rules for exponents.
\(\displaystyle \left( {b^m } \right)^n = \left( {b^n } \right)^m = b^{mn}\).
Thus if \(\displaystyle \left( {\sqrt[n]{a}} \right) = a^\alpha\) what \(\displaystyle \alpha\) would you choose?
\(\displaystyle a = a^1 = \left( {\sqrt[n]{a}} \right)^n = \left( {a^\alpha } \right)^n = a^{\alpha n}\), so it stands to reason that \(\displaystyle \alpha = \frac{1}{n}\).

The fractional exponent makes multiplication much more reasonable.
Using radical notation it is not clear how we would do this: \(\displaystyle \left( {\sqrt[3]{{x^2 }}} \right)\left( {\sqrt[5]{{x^6 }}} \right)\).
But with fractional exponent notation \(\displaystyle \left( {x^{\frac{2}{3}} } \right)\left( {x^{\frac{6}{5}} } \right) = \left( {x^{\frac{{10}}{{15}} + \frac{{18}}{{15}}} } \right) = \left( {x^{\frac{{28}}{{15}}} } \right) = \left( {x^{1 + \frac{{13}}{{15}}} } \right) = x\sqrt[{15}]{{x^{13} }}\)