12 multiple-choice questions, progressively harder.
Evaluate 4−2+404^{-2} + 4^{0}4−2+40 as a fraction.
Solution
Correct answer: A
Evaluate each term, then add over a common denominator.
4−2=116,40=1=16164^{-2} = \frac{1}{16}, \qquad 4^{0} = 1 = \frac{16}{16}4−2=161,40=1=1616
So the sum is
116+1616=1716\frac{1}{16} + \frac{16}{16} = \frac{17}{16}161+1616=1617
Simplify a−3⋅a7a6\dfrac{a^{-3} \cdot a^{7}}{a^{6}}a6a−3⋅a7 to a single power of aaa with a positive exponent (with a≠0a \neq 0a=0).
Add the exponents on top, then subtract the bottom exponent.
a−3⋅a7=a−3+7=a4,a4a6=a4−6=a−2a^{-3} \cdot a^{7} = a^{-3 + 7} = a^{4}, \qquad \frac{a^{4}}{a^{6}} = a^{4-6} = a^{-2}a−3⋅a7=a−3+7=a4,a6a4=a4−6=a−2
Rewriting with a positive exponent gives 1a2\tfrac{1}{a^2}a21.
Evaluate (50+30)−2\left(5^{0} + 3^{0}\right)^{-2}(50+30)−2 as a fraction.
Correct answer: D
Each zero power is 111, so add inside the parentheses first.
50+30=1+1=25^{0} + 3^{0} = 1 + 1 = 250+30=1+1=2
Then apply the outer negative exponent:
2−2=122=142^{-2} = \frac{1}{2^2} = \frac{1}{4}2−2=221=41
Simplify (2−2)2⋅25\left(2^{-2}\right)^{2} \cdot 2^{5}(2−2)2⋅25 to a single power of 222.
Correct answer: C
Resolve the power of a power first, then add the exponents.
(2−2)2=2(−2)(2)=2−4,2−4⋅25=2−4+5=21\left(2^{-2}\right)^{2} = 2^{(-2)(2)} = 2^{-4}, \qquad 2^{-4} \cdot 2^{5} = 2^{-4 + 5} = 2^{1}(2−2)2=2(−2)(2)=2−4,2−4⋅25=2−4+5=21
Which expression is equal to (45)−1\left(\dfrac{4}{5}\right)^{-1}(54)−1?
Correct answer: B
A −1-1−1 exponent inverts the fraction.
(45)−1=54\left(\frac{4}{5}\right)^{-1} = \frac{5}{4}(54)−1=45
Simplify 12 m−1n44 m3n−1\dfrac{12\, m^{-1} n^{4}}{4\, m^{3} n^{-1}}4m3n−112m−1n4 with positive exponents (with m,n≠0m, n \neq 0m,n=0).
Divide the coefficients, then subtract exponents for each base.
124=3,m−1−3=m−4,n4−(−1)=n5\frac{12}{4} = 3, \quad m^{-1-3} = m^{-4}, \quad n^{4 - (-1)} = n^{5}412=3,m−1−3=m−4,n4−(−1)=n5
Moving the negative power of mmm to the denominator gives
3n5m4\frac{3 n^5}{m^4}m43n5
Evaluate (2−1−4−1)−1\left(2^{-1} - 4^{-1}\right)^{-1}(2−1−4−1)−1.
Subtract inside the parentheses first, over a common denominator.
2−1−4−1=12−14=24−14=142^{-1} - 4^{-1} = \frac{1}{2} - \frac{1}{4} = \frac{2}{4} - \frac{1}{4} = \frac{1}{4}2−1−4−1=21−41=42−41=41
Then the outer −1-1−1 exponent inverts it:
(14)−1=4\left(\frac{1}{4}\right)^{-1} = 4(41)−1=4
Order from smallest to largest: A=3−1A = 3^{-1}A=3−1, B=30B = 3^{0}B=30, C=3−2C = 3^{-2}C=3−2.
Evaluate each value.
A=13,B=1,C=19A = \frac{1}{3}, \qquad B = 1, \qquad C = \frac{1}{9}A=31,B=1,C=91
From smallest to largest, 19<13<1\tfrac{1}{9} < \tfrac{1}{3} < 191<31<1, so the order is C,A,BC, A, BC,A,B.
Which expression does NOT equal 888?
Reduce each to a single value.
2−3⋅26=23=8,(12)−3=23=8,2−12−4=23=82^{-3} \cdot 2^{6} = 2^{3} = 8, \quad \left(\frac{1}{2}\right)^{-3} = 2^{3} = 8, \quad \frac{2^{-1}}{2^{-4}} = 2^{3} = 82−3⋅26=23=8,(21)−3=23=8,2−42−1=23=8
But 2−32^{-3}2−3 is a reciprocal:
2−3=182^{-3} = \frac{1}{8}2−3=81
So 2−3=182^{-3} = \tfrac{1}{8}2−3=81 is the one that does not equal 888.
Simplify (a−1a2)−2\left(\dfrac{a^{-1}}{a^{2}}\right)^{-2}(a2a−1)−2 to a single power of aaa (with a≠0a \neq 0a=0).
Simplify inside the parentheses first, then apply the outer power.
a−1a2=a−1−2=a−3,(a−3)−2=a(−3)(−2)=a6\frac{a^{-1}}{a^{2}} = a^{-1 - 2} = a^{-3}, \qquad \left(a^{-3}\right)^{-2} = a^{(-3)(-2)} = a^{6}a2a−1=a−1−2=a−3,(a−3)−2=a(−3)(−2)=a6
If 2x⋅25=222^{x} \cdot 2^{5} = 2^{2}2x⋅25=22, what is xxx?
The product rule adds the exponents, so x+5x + 5x+5 must equal 222.
x+5=2 ⟹ x=−3x + 5 = 2 \implies x = -3x+5=2⟹x=−3
Check: 2−3⋅25=2−3+5=222^{-3} \cdot 2^{5} = 2^{-3 + 5} = 2^{2}2−3⋅25=2−3+5=22.
Evaluate 5−1+2−25^{-1} + 2^{-2}5−1+2−2 as a fraction.
Rewrite each term, then add over a common denominator of 202020.
5−1=15=420,2−2=14=5205^{-1} = \frac{1}{5} = \frac{4}{20}, \qquad 2^{-2} = \frac{1}{4} = \frac{5}{20}5−1=51=204,2−2=41=205
420+520=920\frac{4}{20} + \frac{5}{20} = \frac{9}{20}204+205=209
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