Decimals, roots and identities with exponents

This les­son is a set of shorter prob­lems of the kind that appear in objec­tive tests and short-answer ques­tions. They use the same laws, but each has a twist: a dec­i­mal base, a dec­i­mal expo­nent, a frac­tion with a surd in it, or an alge­braic iden­tity hid­ing behind frac­tional pow­ers. Work through the exam­ples, then try the prac­tice set.

Dec­i­mals as bases#

A dec­i­mal can be writ­ten as a frac­tion with a power of $10$ under­neath, and then often as a power of a sim­ple dec­i­mal.

Exam­ple 1#

Find $(0.001)^{1/3}$.

$$(0.001)^{1/3} = \left[(0.1)^{3}\right]^{1/3} = 0.1$$

Or, using frac­tions: $\displaystyle (0.001)^{1/3} = \left(\frac{1}{1000}\right)^{1/3} = \left[\left(\frac{1}{10}\right)^{3}\right]^{1/3} = \frac{1}{10} = 0.1$.

Exam­ple 2#

Find $\sqrt[5]{32^{-3}}$.

Use Law 13, $\sqrt[n]{a^{m}} = a^{m/n}$.

$$\displaystyle \begin{aligned} \sqrt[5]{32^{-3}} &= 32^{-3/5} \\ &= (2^{5})^{-3/5} \\ &= 2^{-3} = \frac{1}{2^{3}} = \frac{1}{8} \end{aligned}$$

A surd inside the base#

Exam­ple 3#

Find $\displaystyle \left(\frac{\sqrt{2}}{5}\right)^{8} \div \left(\frac{\sqrt{2}}{5}\right)^{13}$.

Use Law 2 first, then Law 8.

$$\displaystyle \begin{aligned} \left(\frac{\sqrt{2}}{5}\right)^{8 - 13} &= \left(\frac{\sqrt{2}}{5}\right)^{-5} \\ &= \left(\frac{5}{\sqrt{2}}\right)^{5} \\ &= \frac{5^{5}}{(\sqrt{2})^{5}} = \frac{3125}{4\sqrt{2}} \end{aligned}$$

Here $5^{5} = 3125$ and $(\sqrt{2})^{5} = (\sqrt{2})^{4} \times \sqrt{2} = 4\sqrt{2}$. Ratio­nal­is­ing the denom­i­na­tor gives $\displaystyle \frac{3125\sqrt{2}}{8}$. A com­mon slip is to write $5^{5}$ as $25$; it is $5 \times 5 \times 5 \times 5 \times 5 = 3125$.

Pow­ers of 10#

Exam­ple 4#

If $5^{3x} = 125$ and $10^{y} = 0.001$, find $x$ and $y$.

For the first, $125 = 5^{3}$, so $5^{3x} = 5^{3}$, giv­ing $3x = 3$ and $x = 1$.

For the sec­ond, $\displaystyle 0.001 = \frac{1}{1000} = 10^{-3}$, so $10^{y} = 10^{-3}$ and $y = -3$. Each dec­i­mal place is one more neg­a­tive power of $10$: $0.1 = 10^{-1}$, $0.01 = 10^{-2}$, $0.001 = 10^{-3}$.

Exam­ple 5#

If $10^{2y} = 25$, find $10^{-y}$.

Write the left side as a square: $(10^{y})^{2} = 5^{2}$, so $10^{y} = 5$ (it must be pos­i­tive). Then

$$\displaystyle 10^{-y} = \frac{1}{10^{y}} = \frac{1}{5}$$

Dec­i­mal expo­nents#

Exam­ple 6#

Find $(256)^{0.16} \times (256)^{0.09}$.

Same base, so add the expo­nents. Then change the dec­i­mal expo­nent to a frac­tion: $\displaystyle 0.25 = \frac{25}{100} = \frac{1}{4}$.

$$\begin{aligned} (256)^{0.16} \times (256)^{0.09} &= (256)^{0.16 + 0.09} \\ &= (256)^{0.25} \\ &= (4^{4})^{1/4} = 4 \end{aligned}$$

Exam­ple 7#

Solve $\displaystyle \left(\frac{2}{3}\right)^{x}\left(\frac{3}{2}\right)^{2x} = \frac{81}{16}$.

Turn the first fac­tor round with Law 8: $\displaystyle \left(\frac{2}{3}\right)^{x} = \left(\frac{3}{2}\right)^{-x}$. Also $\displaystyle \frac{81}{16} = \left(\frac{3}{2}\right)^{4}$.

$$\displaystyle \begin{aligned} \left(\frac{3}{2}\right)^{-x} \cdot \left(\frac{3}{2}\right)^{2x} &= \left(\frac{3}{2}\right)^{4} \\ \left(\frac{3}{2}\right)^{x} &= \left(\frac{3}{2}\right)^{4} \\ x &= 4 \end{aligned}$$

Iden­ti­ties hid­ing behind frac­tional pow­ers#

Frac­tional pow­ers often fit an iden­tity from alge­bra. Two are espe­cially use­ful:

$$(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)$$

$$(a + b)(a^{2} - ab + b^{2}) = a^{3} + b^{3}$$

Exam­ple 8#

If $x = 2^{1/3} + 2^{2/3}$, show that $x^{3} - 6x = 6$.

Cube both sides, using $(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)$ with $a = 2^{1/3}$ and $b = 2^{2/3}$:

$$\begin{aligned} x^{3} &= (2^{1/3})^{3} + (2^{2/3})^{3} + 3 \cdot 2^{1/3} \cdot 2^{2/3}\left(2^{1/3} + 2^{2/3}\right) \\ &= 2 + 2^{2} + 3 \cdot 2 \cdot x \\ &= 2 + 4 + 6x \end{aligned}$$

Here $2^{1/3} \cdot 2^{2/3} = 2^{1} = 2$, and the bracket is $x$ itself. So $x^{3} = 6 + 6x$, that is, $x^{3} - 6x = 6$.

Exam­ple 9#

If $x = 3$, find the value of $\left(x^{1/3} + x^{-1/3}\right)\left(x^{2/3} + x^{-2/3} - 1\right)$.

With $a = x^{1/3}$ and $b = x^{-1/3}$, we have $ab = x^{0} = 1$, so the sec­ond bracket is $a^{2} + b^{2} - ab$. The prod­uct is there­fore $a^{3} + b^{3}$:

$$\displaystyle \begin{aligned} \left(x^{1/3} + x^{-1/3}\right)\left(x^{2/3} + x^{-2/3} - 1\right) &= (x^{1/3})^{3} + (x^{-1/3})^{3} \\ &= x + x^{-1} = x + \frac{1}{x} \end{aligned}$$

For $x = 3$ this is $\displaystyle 3 + \frac{1}{3} = \frac{10}{3}$.

Prac­tice#

1. Find $(0.008)^{1/3}$.
2. Find $(0.0081)^{1/4}$.
3. Find $\sqrt[3]{27^{-2}}$.
4. Find $\displaystyle \left(\frac{3}{4}\right)^{5} \div \left(\frac{3}{4}\right)^{8}$.
5. If $2^{4x} = 16$ and $10^{y} = 0.01$, find $x$ and $y$.
6. Find $(81)^{0.15} \times (81)^{0.10}$.
7. If $10^{3y} = 8$, find $10^{-y}$.
8. Solve $\displaystyle \left(\frac{5}{3}\right)^{x}\left(\frac{3}{5}\right)^{2x} = \frac{27}{125}$.
9. If $x = 3^{1/3} + 3^{-1/3}$, show that $3x^{3} - 9x = 10$.
10. If $x = 2$, find the value of $\left(x^{1/3} - x^{-1/3}\right)\left(x^{2/3} + x^{-2/3} + 1\right)$.

Answers#