Fractions and Exponents Practice: Intermediate

Fractions with letters

Simplify. Every letter stands for a positive number, so no denominator is zero.

1. 1.
   \( \displaystyle \frac{6x^2}{3x} \)
   Solution

   \( \displaystyle = \frac{6}{3} \cdot \frac{x^2}{x} \)
   \( \displaystyle = 2x \)
2. 2.
   \( \displaystyle \frac{x^2 y}{x y^3} \)
   Solution

   \( \displaystyle = \frac{x}{y^2} \)

   Cancel one \( x \) and one \( y \) from top and bottom.

3. 3.
   \( \displaystyle \frac{3x + 6}{3} \)
   Solution

   \( \displaystyle = \frac{3(x + 2)}{3} \)
   \( \displaystyle = x + 2 \)

   Factor the top first, then cancel. Every term on top must be divided, not just the first one.

4. 4.
   \( \displaystyle \frac{2xy + 4x}{2x} \)
   Solution

   \( \displaystyle = \frac{2x(y + 2)}{2x} \)
   \( \displaystyle = y + 2 \)
5. 5.
   \( \displaystyle \frac{x + 3}{x} \)
   Solution

   \( \displaystyle = \frac{x}{x} + \frac{3}{x} \)
   \( \displaystyle = 1 + \frac{3}{x} \)

   A classic trap: this is not 3. You can cancel a factor of the whole top, never a single term of a sum. Splitting the fraction is allowed.

6. 6.
   \( \displaystyle \frac{x^2 - 4}{x - 2} \)
   Solution

   \( \displaystyle = \frac{(x - 2)(x + 2)}{x - 2} \)
   \( \displaystyle = x + 2 \)

   Factor first: \( x^2 - 4 = (x-2)(x+2) \). Then the bracket \( x - 2 \) is a factor and can be cancelled (for \( x \neq 2 \)).

7. 7.
   \( \displaystyle \frac{a}{2} + \frac{a}{3} \)
   Solution

   \( \displaystyle = \frac{3a}{6} + \frac{2a}{6} \)
   \( \displaystyle = \frac{5a}{6} \)
8. 8.
   \( \displaystyle \frac{1}{x} + \frac{1}{y} \)
   Solution

   \( \displaystyle = \frac{y}{xy} + \frac{x}{xy} \)
   \( \displaystyle = \frac{x + y}{xy} \)

   The common denominator of \( x \) and \( y \) is \( xy \). It is not \( \frac{2}{x+y} \).

9. 9.
   \( \displaystyle \frac{2}{x} - \frac{1}{x^2} \)
   Solution

   \( \displaystyle = \frac{2x}{x^2} - \frac{1}{x^2} \)
   \( \displaystyle = \frac{2x - 1}{x^2} \)
10. 10.
    \( \displaystyle \frac{x}{2} - \frac{x - 1}{3} \)
    Solution

    \( \displaystyle = \frac{3x}{6} - \frac{2(x - 1)}{6} \)
    \( \displaystyle = \frac{3x - 2x + 2}{6} \)
    \( \displaystyle = \frac{x + 2}{6} \)

    The minus sign applies to the whole numerator \( 2(x - 1) \), so \( -2(x-1) = -2x + 2 \).

11. 11.
    \( \displaystyle \frac{1}{x} - \frac{1}{x + 1} \)
    Solution

    \( \displaystyle = \frac{x + 1}{x(x + 1)} - \frac{x}{x(x + 1)} \)
    \( \displaystyle = \frac{1}{x(x + 1)} \)
12. 12.
    \( \displaystyle \frac{x}{x + 1} + \frac{1}{x + 1} \)
    Solution

    \( \displaystyle = \frac{x + 1}{x + 1} \)
    \( \displaystyle = 1 \)

    Same denominator already: just add the tops.

13. 13.
    \( \displaystyle 1 - \frac{a}{a + b} \)
    Solution

    \( \displaystyle = \frac{a + b}{a + b} - \frac{a}{a + b} \)
    \( \displaystyle = \frac{b}{a + b} \)
14. 14.
    \( \displaystyle \frac{x}{y} \cdot \frac{y^2}{x^3} \)
    Solution

    \( \displaystyle = \frac{x y^2}{x^3 y} \)
    \( \displaystyle = \frac{y}{x^2} \)
15. 15.
    \( \displaystyle \frac{\alpha}{1 - \alpha} \cdot \frac{1 - \alpha}{\beta} \)
    Solution

    \( \displaystyle = \frac{\alpha}{\beta} \)

    Here \( 1 - \alpha \) is a factor (the whole bracket), so it cancels.

16. 16.
    \( \displaystyle \frac{a/b}{c/d} \)
    Solution

    \( \displaystyle = \frac{a}{b} \cdot \frac{d}{c} \)
    \( \displaystyle = \frac{ad}{bc} \)
17. 17.
    \( \displaystyle \frac{m/p}{m/q} \)
    Solution

    \( \displaystyle = \frac{m}{p} \cdot \frac{q}{m} \)
    \( \displaystyle = \frac{q}{p} \)
18. 18.
    \( \displaystyle \frac{p_1}{p_2} \div \frac{p_1}{2p_2} \)
    Solution

    \( \displaystyle = \frac{p_1}{p_2} \cdot \frac{2p_2}{p_1} \)
    \( \displaystyle = 2 \)
19. 19.
    \( \displaystyle \frac{1/x}{\frac1x + \frac1y} \)
    Solution

    \( \displaystyle = \frac{1/x}{(x + y)/(xy)} \)
    \( \displaystyle = \frac{1}{x} \cdot \frac{xy}{x + y} \)
    \( \displaystyle = \frac{y}{x + y} \)

    Simplify the bottom into one fraction, then divide by flipping it.

Exponent rules with letters

Simplify, and write the answer with positive exponents or as a single power, whichever is asked.

1. 20.
   \( \displaystyle x^3 \cdot x^5 \)
   Solution

   \( \displaystyle = x^{3+5} \)
   \( \displaystyle = x^8 \)
2. 21.
   \( \displaystyle \frac{x^7}{x^3} \)
   Solution

   \( \displaystyle = x^{7-3} \)
   \( \displaystyle = x^4 \)
3. 22.
   \( \displaystyle \left(x^2\right)^4 \)
   Solution

   \( \displaystyle = x^{2 \cdot 4} \)
   \( \displaystyle = x^8 \)
4. 23.
   \( \displaystyle \left(2x^3\right)^2 \)
   Solution

   \( \displaystyle = 2^2 \left(x^3\right)^2 \)
   \( \displaystyle = 4x^6 \)

   The power applies to every factor in the bracket, including the number 2.

5. 24.
   \( \displaystyle \frac{x^2 y^5}{x^4 y^2} \)
   Solution

   \( \displaystyle = x^{2-4} y^{5-2} \)
   \( \displaystyle = x^{-2} y^3 \)
   \( \displaystyle = \frac{y^3}{x^2} \)
6. 25.
   \( \displaystyle \frac{(xy)^3}{x^2 y} \)
   Solution

   \( \displaystyle = \frac{x^3 y^3}{x^2 y} \)
   \( \displaystyle = x y^2 \)
7. 26.
   \( \displaystyle x^{1/2} \cdot x^{1/3} \)
   Solution

   \( \displaystyle = x^{\frac12 + \frac13} \)
   \( \displaystyle = x^{5/6} \)

   Adding the exponents means adding fractions: \( \frac12 + \frac13 = \frac36 + \frac26 = \frac56 \).

8. 27.
   \( \displaystyle \frac{x^{1/2}}{x^{-1/2}} \)
   Solution

   \( \displaystyle = x^{\frac12 - \left(-\frac12\right)} \)
   \( \displaystyle = x \)
9. 28.
   \( \displaystyle \left(x^{2/3}\right)^{3/2} \)
   Solution

   \( \displaystyle = x^{\frac23 \cdot \frac32} \)
   \( \displaystyle = x \)

   Raising to \( \frac32 \) undoes raising to \( \frac23 \).

10. 29.
    \( \displaystyle \left(x^{1/2} y^{1/2}\right)^2 \)
    Solution

    \( \displaystyle = x^{1} y^{1} \)
    \( \displaystyle = xy \)
11. 30.
    \( \displaystyle \left(4x^2\right)^{1/2} \)
    Solution

    \( \displaystyle = 4^{1/2} \left(x^2\right)^{1/2} \)
    \( \displaystyle = 2x \)
12. 31.
    \( \displaystyle \left(8x^3 y^6\right)^{1/3} \)
    Solution

    \( \displaystyle = 8^{1/3} x^{1} y^{2} \)
    \( \displaystyle = 2xy^2 \)
13. 32.
    \( \displaystyle \frac{1}{x^3} \text{ as a single power} \)
    Solution

    \( \displaystyle = x^{-3} \)
14. 33.
    \( \displaystyle \frac{1}{\sqrt{x}} \text{ as a single power} \)
    Solution

    \( \displaystyle = \frac{1}{x^{1/2}} \)
    \( \displaystyle = x^{-1/2} \)

    Rewriting roots and fractions as powers like this is the first step of almost every derivative on the next sheet.

15. 34.
    \( \displaystyle x\sqrt{x} \text{ as a single power} \)
    Solution

    \( \displaystyle = x^1 \cdot x^{1/2} \)
    \( \displaystyle = x^{3/2} \)
16. 35.
    \( \displaystyle \frac{3}{x^2} \text{ as a number times a power} \)
    Solution

    \( \displaystyle = 3x^{-2} \)

    Only \( x^2 \) moves; the 3 was already on top and stays there.

17. 36.
    \( \displaystyle \frac{1}{2x} \text{ as a number times a power} \)
    Solution

    \( \displaystyle = \frac{1}{2} \cdot \frac{1}{x} \)
    \( \displaystyle = \frac{1}{2}x^{-1} \)

    Careful: \( \frac{1}{2x} \) is \( \frac12 x^{-1} \), not \( 2x^{-1} \).

18. 37.
    \( \displaystyle \frac{\sqrt[3]{x^2}}{x} \text{ as a single power} \)
    Solution

    \( \displaystyle = \frac{x^{2/3}}{x^1} \)
    \( \displaystyle = x^{-1/3} \)
19. 38.
    \( \displaystyle (tx)^{a}\,(ty)^{1-a} \)
    Solution

    \( \displaystyle = t^{a} x^{a}\, t^{1-a} y^{1-a} \)
    \( \displaystyle = t^{a + 1 - a} x^{a} y^{1-a} \)
    \( \displaystyle = t\, x^{a} y^{1-a} \)

    Doubling both inputs (\( t = 2 \)) doubles the output: this Cobb-Douglas function has constant returns to scale.

20. 39.
    \( \displaystyle \frac{\tfrac13 x^{-2/3} y^{2/3}}{\tfrac23 x^{1/3} y^{-1/3}} \)
    Solution

    \( \displaystyle = \frac{1/3}{2/3} \cdot x^{-\frac23 - \frac13}\, y^{\frac23 + \frac13} \)
    \( \displaystyle = \frac12\, x^{-1} y \)
    \( \displaystyle = \frac{y}{2x} \)

    This is the MRS of \( U = x^{1/3} y^{2/3} \). Numbers with numbers, \( x \) with \( x \), \( y \) with \( y \).

21. 40.
    \( \displaystyle \frac{a\, x^{a-1} y^{b}}{b\, x^{a} y^{b-1}} \)
    Solution

    \( \displaystyle = \frac{a}{b}\, x^{(a-1) - a}\, y^{b - (b-1)} \)
    \( \displaystyle = \frac{a}{b}\, x^{-1} y \)
    \( \displaystyle = \frac{a\,y}{b\,x} \)

    The general Cobb-Douglas MRS. Subtracting exponents that contain letters: \( (a - 1) - a = -1 \) and \( b - (b - 1) = 1 \).

Solving equations

Solve for the letter shown. Letters are positive unless the equation says otherwise.

1. 41.
   \( \displaystyle \frac{x}{3} + \frac{x}{6} = 5 \)solve for \( x \)
   Solution

   \( \displaystyle 2x + x = 30 \)
   \( \displaystyle \Rightarrow\; 3x = 30 \)
   \( \displaystyle \Rightarrow\; x = 10 \)

   Multiply every term by the common denominator 6 to clear all the fractions at once.

2. 42.
   \( \displaystyle \frac{2}{x} = \frac{1}{4} \)solve for \( x \)
   Solution

   \( \displaystyle 2 \cdot 4 = 1 \cdot x \)
   \( \displaystyle \Rightarrow\; x = 8 \)

   Cross-multiply: \( \frac{a}{b} = \frac{c}{d} \) becomes \( ad = bc \).

3. 43.
   \( \displaystyle \frac{3}{x + 1} = 1 \)solve for \( x \)
   Solution

   \( \displaystyle 3 = x + 1 \)
   \( \displaystyle \Rightarrow\; x = 2 \)
4. 44.
   \( \displaystyle \frac{x - 1}{2} = \frac{x + 2}{5} \)solve for \( x \)
   Solution

   \( \displaystyle 5(x - 1) = 2(x + 2) \)
   \( \displaystyle \Rightarrow\; 5x - 5 = 2x + 4 \)
   \( \displaystyle \Rightarrow\; 3x = 9 \)
   \( \displaystyle \Rightarrow\; x = 3 \)

   Brackets matter when you cross-multiply: the whole numerator gets multiplied.

5. 45.
   \( \displaystyle \frac{10}{1 + r} = 8 \)solve for \( r \)
   Solution

   \( \displaystyle 10 = 8(1 + r) \)
   \( \displaystyle \Rightarrow\; 1 + r = \frac{10}{8} = \frac54 \)
   \( \displaystyle \Rightarrow\; r = \frac14 \)

   A present-value equation: 10 euros next year are worth 8 today when the interest rate is 25%.

6. 46.
   \( \displaystyle x^2 = 49 \)solve for \( x \)
   Solution

   \( \displaystyle x = 49^{1/2} \)
   \( \displaystyle \Rightarrow\; x = 7 \)

   With \( x > 0 \) only the positive root counts.

7. 47.
   \( \displaystyle x^{1/2} = 6 \)solve for \( x \)
   Solution

   \( \displaystyle \left(x^{1/2}\right)^2 = 6^2 \)
   \( \displaystyle \Rightarrow\; x = 36 \)

   To undo an exponent, raise both sides to the reciprocal exponent: the reciprocal of \( \frac12 \) is 2.

8. 48.
   \( \displaystyle x^{1/3} = 2 \)solve for \( x \)
   Solution

   \( \displaystyle x = 2^3 \)
   \( \displaystyle \Rightarrow\; x = 8 \)
9. 49.
   \( \displaystyle x^{-1} = 4 \)solve for \( x \)
   Solution

   \( \displaystyle \frac{1}{x} = 4 \)
   \( \displaystyle \Rightarrow\; x = \frac14 \)
10. 50.
    \( \displaystyle 2x^3 = 54 \)solve for \( x \)
    Solution

    \( \displaystyle x^3 = 27 \)
    \( \displaystyle \Rightarrow\; x = 27^{1/3} \)
    \( \displaystyle \Rightarrow\; x = 3 \)

    Divide by 2 first, then undo the cube.

11. 51.
    \( \displaystyle x^{-1/2} = \frac{1}{5} \)solve for \( x \)
    Solution

    \( \displaystyle \frac{1}{x^{1/2}} = \frac15 \)
    \( \displaystyle \Rightarrow\; x^{1/2} = 5 \)
    \( \displaystyle \Rightarrow\; x = 25 \)
12. 52.
    \( \displaystyle \tfrac{1}{2}\, x^{-1/2} = \tfrac{1}{4} \)solve for \( x \)
    Solution

    \( \displaystyle x^{-1/2} = \frac12 \)
    \( \displaystyle \Rightarrow\; x^{1/2} = 2 \)
    \( \displaystyle \Rightarrow\; x = 4 \)

    This is what a first-order condition looks like: the derivative of \( \sqrt x \) set equal to a price of \( \frac14 \).

13. 53.
    \( \displaystyle x^{2/3} = 9 \)solve for \( x \)
    Solution

    \( \displaystyle x = 9^{3/2} \)
    \( \displaystyle \Rightarrow\; x = \left(\sqrt 9\right)^3 = 27 \)

    Raise both sides to \( \frac32 \), the reciprocal of \( \frac23 \).

14. 54.
    \( \displaystyle 2^x = 32 \)solve for \( x \)
    Solution

    \( \displaystyle 2^x = 2^5 \)
    \( \displaystyle \Rightarrow\; x = 5 \)

    When the unknown is in the exponent, write both sides as powers of the same base (or take logs: \( x = \frac{\ln 32}{\ln 2} \)).

15. 55.
    \( \displaystyle \frac{y}{2x} = \frac34 \)solve for \( y \)
    Solution

    \( \displaystyle y = \frac34 \cdot 2x \)
    \( \displaystyle \Rightarrow\; y = \frac{3x}{2} \)

    Solving for \( y \) in terms of \( x \): \( x \) is treated like a number. This is MRS = price ratio in a consumer problem.

16. 56.
    \( \displaystyle x^{1/2} y^{1/2} = 10 \)solve for \( y \)
    Solution

    \( \displaystyle y^{1/2} = \frac{10}{x^{1/2}} \)
    \( \displaystyle \Rightarrow\; y = \frac{100}{x} \)

    The indifference curve at utility 10, written as \( y \) in terms of \( x \). Square both sides: \( \left(\frac{10}{x^{1/2}}\right)^2 = \frac{100}{x} \).

17. 57.
    \( \displaystyle 3x + 2y = 12 \text{ with } y = \tfrac32 x \)solve for \( x \)
    Solution

    \( \displaystyle 3x + 2 \cdot \frac{3x}{2} = 12 \)
    \( \displaystyle \Rightarrow\; 6x = 12 \)
    \( \displaystyle \Rightarrow\; x = 2 \text{ and } y = 3 \)

    Substitute the expression for \( y \) into the budget line. The 2 cancels with the denominator 2.

18. 58.
    \( \displaystyle p_x x = \frac{a}{1 - a}\, p_y y \text{ and } p_x x + p_y y = m \)solve for \( p_y y \)
    Solution

    \( \displaystyle \frac{a}{1 - a}\,p_y y + p_y y = m \)
    \( \displaystyle \Rightarrow\; \frac{a + 1 - a}{1 - a}\, p_y y = m \)
    \( \displaystyle \Rightarrow\; p_y y = (1 - a)\, m \)

    Solve for the spending \( p_y y \) as one block. Then \( y = \frac{(1-a)m}{p_y} \) and \( x = \frac{a m}{p_x} \): the Cobb-Douglas demands.