1
$\begingroup$

how would you explain the difference between exponential multiplication and fractional multiplication? $${x^{1/3}}{^{}{}} * {x^{1/3}}{^{}{}} *{x^{1/3}}{^{}{}} * {x^{1/3}}{^{}{}} = {x^{4/3}}$$ Why is this the same as $$4 * {^{1/3}}$$ On the other hand $$1/3 * 1/3 * 1/3 * 1/3 = 1/81$$ So in this case, this is not the same as $$4 * {{1/3}}$$

$\endgroup$
  • 1
    $\begingroup$ The property says $$x^\alpha\cdot x^\beta=x^{\alpha+\beta}$$ Exponents product happens with powers of powers, like $$\left(x^\alpha\right)^{\beta}=x^{\alpha\beta}$$ $\endgroup$ – Raffaele Nov 25 '17 at 17:38
1
$\begingroup$

Depending on the level of student, the explanation might go as follows:

First, note that if $m$ and $n$ are integers, then we might regard exponentiation as repeated multiplication $$ x^m \cdot x^n = (\underbrace{x\cdot x\cdot \dotsb \cdot x}_{\text{$m$ times}})\cdot (\underbrace{x\cdot x\cdot \dotsb \cdot x}_{\text{$n$ times}}) = \underbrace{x\cdot x\cdot \dotsb \cdot x}_{\text{$m+n$ times}} = x^{m+n}. $$ This is a little bit of a lie (or not—again, it depends on the mathematical maturity of the audience). This can be worked out explicitly with small $m$ and $n$, and/or with fixed values of $x$ (say, work out $2^3\cdot 2^5$ by hand).

Then, when fractional exponents are introduced, it is reasonable to extend the operation above: if $p$ and $q$ are rational, then we should have $$ x^p\cdot x^q = x^{p+q}. $$ Again, with small $x$, $p$, and $q$, this could be justified by hand. Consider $$ 3^{\frac{1}{2}} \cdot 3^{\frac{5}{2}} = \sqrt{3}^1 \cdot \sqrt{3}^5 = \sqrt{3}^{5+1} = 3^{\frac{5+1}{2}} = 3^{\frac{5}{2} + \frac{1}{2}}. $$ This isn't really rigorous, but it should make the idea clear to students how have not seen it before. The moral of the story is that when terms with like bases are multiplied, the exponents are added.

This is in contrast to something like $$ \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} = \left( \frac{1}{3} \right)^4 = \frac{1}{3^4} = \frac{1}{81}. $$ Here we are again using the idea that exponentiation can be viewed as repeated multiplication. Compare this to something like $$ 3+3+3+3 = 4\cdot 3,$$ which is not the same as $$ 3+3+3+3 = 4+3. $$ Writing the second identity would be the same kind of error as writing $$ \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} = 4 \cdot \frac{1}{3}. $$

$\endgroup$
  • $\begingroup$ Thank you for taking the time Xander, much appreciated. $\endgroup$ – Lani Nov 25 '17 at 18:18
2
$\begingroup$

First, remember that $$b^x\cdot b^y=b^{x+y}.$$ That’s why $$x^{1/3}\cdot x^{1/3}\cdot x^{1/3}\cdot x^{1/3}=x^{1/3+1/3+1/3+1/3}=x^{4\cdot 1/3}=x^{4/3}.$$ One the other hand $$\frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}=\frac{1\cdot 1\cdot 1\cdot 1}{3\cdot 3\cdot 3\cdot 3}=\frac{1}{81}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.