12 multiple-choice questions, progressively harder.
Simplify (x4x)2\left(\dfrac{x^4}{x}\right)^2(xx4)2 to a single power of xxx (with x≠0x \neq 0x=0).
Solution
Correct answer: A
Simplify inside the parentheses with the quotient rule, then apply the outer power.
x4x=x4−1=x3,(x3)2=x3×2=x6\frac{x^4}{x} = x^{4-1} = x^3, \qquad (x^3)^2 = x^{3 \times 2} = x^6xx4=x4−1=x3,(x3)2=x3×2=x6
Evaluate (22)3(2^2)^3(22)3 as a single whole number.
Correct answer: B
Multiply the exponents, then evaluate the power since a number is requested.
(22)3=26=64(2^2)^3 = 2^6 = 64(22)3=26=64
Simplify (32×34)237\dfrac{(3^2 \times 3^4)^2}{3^7}37(32×34)2 to a single power of 333.
Correct answer: D
Combine inside the parentheses, apply the outer power, then divide.
32×34=36,(36)2=312,31237=312−7=353^2 \times 3^4 = 3^6, \qquad (3^6)^2 = 3^{12}, \qquad \frac{3^{12}}{3^7} = 3^{12-7} = 3^532×34=36,(36)2=312,37312=312−7=35
Order from largest to smallest: A=(22)4A = (2^2)^4A=(22)4, B=25×22B = 2^5 \times 2^2B=25×22, C=21223C = \dfrac{2^{12}}{2^3}C=23212.
Correct answer: C
Reduce each to a single power of 222 and compare exponents.
A=28,B=27,C=29A = 2^8, \qquad B = 2^7, \qquad C = 2^9A=28,B=27,C=29
Since 9>8>79 > 8 > 79>8>7, the order largest to smallest is C,A,BC, A, BC,A,B.
Which statement is true?
Use the power-of-a-product rule in reverse: 23×53=(2×5)3=1032^3 \times 5^3 = (2 \times 5)^3 = 10^323×53=(2×5)3=103.
23×53=(2×5)3=1032^3 \times 5^3 = (2 \times 5)^3 = 10^323×53=(2×5)3=103
The others fail: the exponent on 101010 stays 333 (not 666); 23+23=2×23=242^3 + 2^3 = 2 \times 2^3 = 2^423+23=2×23=24; and a power does not split over a sum.
Fill in the blank so the statement is true: (53)□=56×56(5^3)^{\square} = 5^6 \times 5^6(53)□=56×56.
Reduce both sides to a single power of 555. The right side adds exponents; the left multiplies.
56×56=512,(53)□=53□ ⇒ 3□=12 ⇒ □=45^6 \times 5^6 = 5^{12}, \qquad (5^3)^{\square} = 5^{3\square} \;\Rightarrow\; 3\square = 12 \;\Rightarrow\; \square = 456×56=512,(53)□=53□⇒3□=12⇒□=4
Simplify (a3)2×aa4\dfrac{(a^3)^2 \times a}{a^4}a4(a3)2×a to a single power of aaa (with a≠0a \neq 0a=0).
Resolve the power of a power, add the lone factor's exponent on top, then subtract the bottom.
(a3)2=a6,a6×a1=a7,a7a4=a7−4=a3(a^3)^2 = a^6, \qquad a^6 \times a^1 = a^7, \qquad \frac{a^7}{a^4} = a^{7-4} = a^3(a3)2=a6,a6×a1=a7,a4a7=a7−4=a3
Simplify (32)3×334\dfrac{(3^2)^3 \times 3}{3^4}34(32)3×3 to a single power of 333.
Resolve the power of a power, add the lone factor's exponent (3=313 = 3^13=31), then subtract the bottom.
(32)3=36,36×31=37,3734=37−4=33(3^2)^3 = 3^6, \qquad 3^6 \times 3^1 = 3^7, \qquad \frac{3^7}{3^4} = 3^{7-4} = 3^3(32)3=36,36×31=37,3437=37−4=33
Fill in the blank: 2□×24=(23)22^{\square} \times 2^4 = (2^3)^22□×24=(23)2.
Reduce the right side first, then match exponents using the product rule on the left.
(23)2=26,□+4=6 ⇒ □=2(2^3)^2 = 2^6, \qquad \square + 4 = 6 \;\Rightarrow\; \square = 2(23)2=26,□+4=6⇒□=2
Check: 22×24=262^2 \times 2^4 = 2^622×24=26.
Simplify 8425\dfrac{8^4}{2^5}2584 to a single power of 222. (Hint: 8=238 = 2^38=23.)
Rewrite 888 as 232^323 so the bases match, then subtract the exponents.
84=(23)4=212,21225=212−5=278^4 = (2^3)^4 = 2^{12}, \qquad \frac{2^{12}}{2^5} = 2^{12-5} = 2^784=(23)4=212,25212=212−5=27
Simplify 12545\dfrac{12^5}{4^5}45125 to a single power of 333. (Hint: 124=3\tfrac{12}{4} = 3412=3.)
Both powers share the exponent 555, so use the power-of-a-quotient rule in reverse.
12545=(124)5=35\frac{12^5}{4^5} = \left(\frac{12}{4}\right)^5 = 3^545125=(412)5=35
Simplify (22×3)3(2^2 \times 3)^3(22×3)3 to the form 2a×3b2^a \times 3^b2a×3b.
Give the outer exponent to each factor, then resolve the power of a power on the 222^222.
(22×3)3=(22)3×33=26×33(2^2 \times 3)^3 = (2^2)^3 \times 3^3 = 2^6 \times 3^3(22×3)3=(22)3×33=26×33
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