13. Marked exercises (25 Sep) — Fractions

27. Powers with the same base or exponent

Write the following operations with powers as a single power and work out its value with the help of a calculator.

a) $\displaystyle \left(-\dfrac{3}{2}\right)^5 \cdot \left(-\dfrac{3}{2}\right)^2 : \left(-\dfrac{3}{2}\right)^3$

Same base: add the exponents when multiplying and subtract them when dividing.

\left(-\dfrac{3}{2}\right)^{5+2-3} = \left(-\dfrac{3}{2}\right)^4 = \dfrac{3^4}{2^4} = \dfrac{81}{16}

The exponent is even, so the minus sign disappears.

\left(-\dfrac{3}{2}\right)^4 = \dfrac{81}{16}

c) $\displaystyle \left[\left(\dfrac{4}{5}\right)^2\right]^3 : \left(\dfrac{4}{5}\right)^2 \cdot \left(\dfrac{10}{3}\right)^4$

Power of a power: multiply the exponents.

\left(\dfrac{4}{5}\right)^6 : \left(\dfrac{4}{5}\right)^2 \cdot \left(\dfrac{10}{3}\right)^4 = \left(\dfrac{4}{5}\right)^4 \cdot \left(\dfrac{10}{3}\right)^4

Now they share the exponent, so multiply the bases.

\left(\dfrac{4}{5} \cdot \dfrac{10}{3}\right)^4 = \left(\dfrac{40}{15}\right)^4 = \left(\dfrac{8}{3}\right)^4 = \dfrac{4096}{81}

\left(\dfrac{8}{3}\right)^4 = \dfrac{4096}{81}

29. Root of a fraction

Work out the roots of fractions below, using the definition of the root of a fraction. Give the result as an irreducible fraction.

b) $\displaystyle \sqrt{\dfrac{484}{784}}$

The root of a fraction is the root of the numerator over the root of the denominator. Factorise both numbers:

484 = 2^2 \cdot 11^2 \qquad 784 = 2^4 \cdot 7^2

Each pair of equal factors comes out of the root as a single factor:

\dfrac{\sqrt{484}}{\sqrt{784}} = \dfrac{\sqrt{2^2 \cdot 11^2}}{\sqrt{2^4 \cdot 7^2}} = \dfrac{2 \cdot 11}{2^2 \cdot 7} = \dfrac{22}{28} = \dfrac{11}{14}

\dfrac{11}{14}

34. Combined operations with powers and roots

Carry out these combined operations with fractions and simplify the result if possible.

a) $\displaystyle \left(\dfrac{6}{5}\right)^5 : \left(\dfrac{6}{5}\right)^3 - \sqrt{\dfrac{15}{6}} \cdot \sqrt{\dfrac{49}{40}} + \left[\left(\dfrac{5}{4}\right)^2\right]^{-1}$

Both roots have the same index, so they can be merged into one.

\sqrt{\dfrac{15}{6}} \cdot \sqrt{\dfrac{49}{40}} = \sqrt{\dfrac{15 \cdot 49}{6 \cdot 40}} = \sqrt{\dfrac{49}{16}} = \dfrac{7}{4}

The powers: same base in the first one, power of a power in the last one.

\left(\dfrac{6}{5}\right)^{5-3} - \dfrac{7}{4} + \left(\dfrac{5}{4}\right)^{-2} = \dfrac{36}{25} - \dfrac{7}{4} + \dfrac{16}{25}

$\text{LCM}(25, 4) = 100$:

\dfrac{144}{100} - \dfrac{175}{100} + \dfrac{64}{100} = \dfrac{33}{100}

\dfrac{33}{100}

b) $\displaystyle \dfrac{3}{4} : 5 + \left(\dfrac{7}{2}\right)^3 : \left(\dfrac{14}{6}\right)^3 - \left(\dfrac{1}{5}\right)^2 \cdot \left(\dfrac{1}{5}\right)^{-3}$

Powers first. The two in the middle share the exponent; the last two share the base:

\left(\dfrac{7}{2} : \dfrac{14}{6}\right)^3 = \left(\dfrac{42}{28}\right)^3 = \left(\dfrac{3}{2}\right)^3 = \dfrac{27}{8} \qquad \left(\dfrac{1}{5}\right)^{2-3} = \left(\dfrac{1}{5}\right)^{-1} = 5

The division: $\dfrac{3}{4} : 5 = \dfrac{3}{4} : \dfrac{5}{1} = \dfrac{3}{20}$. We are left with:

\dfrac{3}{20} + \dfrac{27}{8} - 5 = \dfrac{6}{40} + \dfrac{135}{40} - \dfrac{200}{40} = -\dfrac{59}{40}

-\dfrac{59}{40}

d) $\displaystyle \sqrt{\dfrac{9}{121}} + \dfrac{5}{12} \cdot \dfrac{24}{25} - \dfrac{1}{2} : \left(-\dfrac{2}{3} + \dfrac{1}{4}\right)$

The root, the multiplication (simplifying first) and the bracket:

\sqrt{\dfrac{9}{121}} = \dfrac{3}{11} \qquad \dfrac{5}{12} \cdot \dfrac{24}{25} = \dfrac{5 \cdot 24}{12 \cdot 25} = \dfrac{2}{5} \qquad -\dfrac{2}{3} + \dfrac{1}{4} = -\dfrac{8}{12} + \dfrac{3}{12} = -\dfrac{5}{12}

The division, crosswise. Minus times minus is plus:

-\dfrac{1}{2} : \left(-\dfrac{5}{12}\right) = +\dfrac{12}{10} = \dfrac{6}{5}

$\text{LCM}(11, 5) = 55$:

\dfrac{3}{11} + \dfrac{2}{5} + \dfrac{6}{5} = \dfrac{15}{55} + \dfrac{22}{55} + \dfrac{66}{55} = \dfrac{103}{55}

\dfrac{103}{55}