Exercises - Powers, Exponents, and their Dependencies

  1. Assuming all variables are real values, simplify each completely (remember the convention of using only positive exponents in your answer):

    1. $\displaystyle{(-4a^{-2}b^4)^2(-a^2b^{-4}c)^3(ab^{-3}c^4)^2}$

    2. $\displaystyle{\frac{18xy^3z^{-5}}{27x^{-2}y^2z^2}}$

    3. $\displaystyle{(7a^2 b^4)(-2a^{-4} b^3)}$

    4. $\displaystyle{(4x^4 y^{-2})(-3x^5 y^{-4})}$

    5. $\displaystyle{\frac{24pq^8 r^{-4}}{36 p^{-3} q^{-6} r^5}}$

    6. $\displaystyle{\frac{54 x^6 y^{-4} z^2}{9x^{-3} y^2 z^{-4}}}$

    7. $\displaystyle{\frac{(2k^2 rt)^{-2}}{(3k^{-1} r^{-2} t^3 )^{-1}}}$

    8. $\displaystyle{\left( \frac{45x^{-2} y^3 (zw^{-1})^0}{(-3)^3 x^{-4} y^{-2} w z^3} \right)^{-2}\!\!\!\!\cdot y^3}$

    1. ${\displaystyle{\frac{-16a^4c^{11}}{b^{10}}}}$

    2. ${\displaystyle{\frac{2x^3y}{3z^7}}}$

    3. ${\displaystyle{\frac{-14b^7}{a^2}}}$

    4. ${\displaystyle{-\frac{12x^9}{y^6}}}$

    5. ${\displaystyle{\frac{2p^4q^{14}}{3r^9}}}$

    6. ${\displaystyle{\frac{6x^9z^6}{y^6}}}$

    7. ${\displaystyle{\frac{3t}{4k^5r^4}}}$

    8. ${\displaystyle{\frac{9 w^2 z^6}{25 x^4 y^7}}}$

  2. Using only the associative property of multiplication [i.e., $(xy)z = x(yz)$], prove that $x^3 \cdot x^6 = x^4 \cdot x^5$.

    Hint: use an argument similar to the one in the notes where $x^4 \cdot x^4$ was shown to be equivalent to $x^5 \cdot x^3$

    $$\begin{array}{rcl} x^3 \cdot x^6 &=& ((x \cdot x) \cdot x) \cdot (((((x \cdot x) \cdot x ) \cdot x) \cdot x) \cdot x)\\ &=& ((x \cdot x) \cdot x) \cdot ((((x \cdot (x \cdot x)) \cdot x) \cdot x) \cdot x)\\ &=& ((x \cdot x) \cdot x) \cdot (((x \cdot ((x \cdot x) \cdot x)) \cdot x) \cdot x)\\ &=& ((x \cdot x) \cdot x) \cdot ((x \cdot (((x \cdot x) \cdot x) \cdot x)) \cdot x)\\ &=& ((x \cdot x) \cdot x) \cdot (x \cdot ((((x \cdot x) \cdot x) \cdot x) \cdot x))\\ &=& (((x \cdot x) \cdot x) \cdot x ) \cdot ((((x \cdot x) \cdot x) \cdot x) \cdot x)\\ &=& x^4 \cdot x^5 \end{array}$$

  3. Using only commutativity and general associativity (i.e., you can regroup a product of $n$ factors any way you wish) prove that $(xyz)^2 = x^2 y^2 z^2$.

    Hint: use an argument similar to the one in the notes that proved $(xy)^3 = x^3 y^3$

    $$\begin{array}{rcl} (xyz)^2 &=& (xyz)(xyz)\\ &=& xy(xz)yz\\ &=& x(xy)zyz\\ &=& xxy(yz)z\\ &=& x^2 y^2 z^2 \end{array}$$

  4. As you have seen, a power of $x$ is the result of repeated multiplication by $x$ some number of times. In particular, $x^m = \overbrace{x \cdot x \cdot x \cdot x \cdots x \cdot x}^{m \textrm{ times}}$ where the individual products are calculated "left-to-right" -- in keeping with the normal rules for "order of operations".

    Putting parentheses in the equation above to make this "left-to-right" order explicit gives us $$x^m = ((((((x \cdot x) \cdot x) \cdot x) \cdots ) \cdot x) \cdot x)$$

    Now consider a power tower, like the following: $$x^{x^{x^{x^{x^{x^{x^{x}}}}}}}$$ Let us denote this more efficiently by $x \uparrow \uparrow 8$, since their are eight $x$'s combined via exponentiation. $x \uparrow \uparrow m$ would be defined similarly.

    Typing powers in one's calculator or computer frequently requires the use of the "caret" operator (e.g., $x\wedge3$ as an alternative to $x^3$). Which means we could write $$x \uparrow\uparrow 8 = x \wedge x \wedge x \wedge x \wedge x \wedge x \wedge x \wedge x$$ Interestingly, the standard way to calculate expressions like the one immediately above is to perform the related individual exponentiations in "right-to-left" order. That is to say, $$x \uparrow\uparrow m = ( x \wedge ( x \wedge ( \cdots ( x \wedge ( x \wedge (x \wedge x))))))$$ With this in mind, answer the following questions:

    1. Calculate the value of $2 \uparrow\uparrow 3$

    2. Calculate the value of $2 \uparrow\uparrow 4$

    3. Calculate $(((2 \wedge 2) \wedge 2) \wedge 2)$

    4. Calculate $2^8$

    5. Why should the exponentiations, when written using carets, be calculated in "right-to-left" order? Keep in mind this was a choice -- mathematicians could have gone the other way and proclaimed such things be done "left-to-right". Why didn't they?

    1. $16$

    2. $65536$

    3. $256$

    4. $256$

    5. Defining $x \uparrow\uparrow m$ to be done "left-to-right" would result in the same value as $2^{2^{m-1}}$, which is something for which we already have a fairly efficient notation. Doing things "right-to-left" however, results in something different -- something new!