12 multiple-choice questions, progressively harder.
Simplify (23)4×2226\dfrac{(2^3)^4 \times 2^2}{2^6}26(23)4×22 to a single power of 222.
Solution
Correct answer: B
Resolve the power of a power, then add on top, then subtract the bottom.
(23)4=212,212×22=214,21426=214−6=28(2^3)^4 = 2^{12}, \qquad 2^{12} \times 2^2 = 2^{14}, \qquad \frac{2^{14}}{2^6} = 2^{14-6} = 2^8(23)4=212,212×22=214,26214=214−6=28
Fill in the blank: (3□)2=310(3^{\square})^2 = 3^{10}(3□)2=310.
Correct answer: D
The power rule multiplies the exponents, so the missing exponent times 222 is 101010.
□×2=10 ⇒ □=5\square \times 2 = 10 \;\Rightarrow\; \square = 5□×2=10⇒□=5
Check: (35)2=310(3^5)^2 = 3^{10}(35)2=310.
Simplify 6723×33\dfrac{6^7}{2^3 \times 3^3}23×3367 to a single power of 666.
Correct answer: A
Rewrite the denominator as a power of 666, since 6=2×36 = 2 \times 36=2×3 gives 23×33=632^3 \times 3^3 = 6^323×33=63, then divide.
6723×33=6763=67−3=64\frac{6^7}{2^3 \times 3^3} = \frac{6^7}{6^3} = 6^{7-3} = 6^423×3367=6367=67−3=64
Which expression is NOT equal to 2122^{12}212?
Correct answer: C
Check each with one law. The power rule gives (23)4=212(2^3)^4 = 2^{12}(23)4=212; the product rule gives 27×25=2122^7 \times 2^5 = 2^{12}27×25=212; the quotient rule gives 21523=212\tfrac{2^{15}}{2^3} = 2^{12}23215=212. But a sum has no exponent law.
24+28=16+256=272≠4096=2122^4 + 2^8 = 16 + 256 = 272 \neq 4096 = 2^{12}24+28=16+256=272=4096=212
So 24+282^4 + 2^824+28 is the one that does not equal 2122^{12}212.
Simplify a5×a4a2×a3\dfrac{a^5 \times a^4}{a^2 \times a^3}a2×a3a5×a4 to a single power of aaa (with a≠0a \neq 0a=0).
Combine the top and the bottom separately with the product rule, then subtract.
a9a5=a9−5=a4\frac{a^9}{a^5} = a^{9-5} = a^4a5a9=a9−5=a4
The top is a5+4=a9a^{5+4} = a^9a5+4=a9 and the bottom is a2+3=a5a^{2+3} = a^5a2+3=a5.
Order from smallest to largest: A=23×23A = 2^3 \times 2^3A=23×23, B=(23)3B = (2^3)^3B=(23)3, C=23×24C = 2^3 \times 2^4C=23×24.
Reduce each to a single power of 222 and compare the exponents.
A=26,B=29,C=27A = 2^6, \qquad B = 2^9, \qquad C = 2^7A=26,B=29,C=27
Since 6<7<96 < 7 < 96<7<9, the order smallest to largest is A,C,BA, C, BA,C,B.
Fill in the blank: 595□=54\dfrac{5^9}{5^{\square}} = 5^45□59=54.
The quotient rule subtracts the exponents, so 999 minus the missing exponent is 444.
9−□=4 ⇒ □=59 - \square = 4 \;\Rightarrow\; \square = 59−□=4⇒□=5
Check: 5955=54\tfrac{5^9}{5^5} = 5^45559=54.
Which single power of 121212 equals 26×332^6 \times 3^326×33? (Hint: 12=22×312 = 2^2 \times 312=22×3.)
Group the factors to match 12=22×312 = 2^2 \times 312=22×3. There are three groups of 22×32^2 \times 322×3, since 26=(22)32^6 = (2^2)^326=(22)3 and 33=3×3×33^3 = 3 \times 3 \times 333=3×3×3.
26×33=(22)3×33=(22×3)3=1232^6 \times 3^3 = (2^2)^3 \times 3^3 = (2^2 \times 3)^3 = 12^326×33=(22)3×33=(22×3)3=123
Simplify 43×254^3 \times 2^543×25 to a single power of 222. (Hint: 4=224 = 2^24=22.)
Rewrite the base 444 as 222^222 so both factors share the base 222, then combine.
43=(22)3=26,26×25=26+5=2114^3 = (2^2)^3 = 2^6, \qquad 2^6 \times 2^5 = 2^{6+5} = 2^{11}43=(22)3=26,26×25=26+5=211
Evaluate 35×3234\dfrac{3^5 \times 3^2}{3^4}3435×32 as a single whole number.
Add the exponents on top, subtract the bottom, then evaluate the power since a number is requested.
3734=37−4=33=27\frac{3^7}{3^4} = 3^{7-4} = 3^3 = 273437=37−4=33=27
If 2n×23=2102^n \times 2^3 = 2^{10}2n×23=210, what is the value of nnn?
The product rule adds the exponents, so n+3=10n + 3 = 10n+3=10.
n+3=10 ⇒ n=7n + 3 = 10 \;\Rightarrow\; n = 7n+3=10⇒n=7
Check: 27×23=2102^7 \times 2^3 = 2^{10}27×23=210.
Simplify a6×b4a2×b\dfrac{a^6 \times b^4}{a^2 \times b}a2×ba6×b4 (with a,b≠0a, b \neq 0a,b=0).
Apply the quotient rule to each base separately, since powers of different bases stay apart.
a6a2=a4,b4b1=b3\frac{a^6}{a^2} = a^4, \qquad \frac{b^4}{b^1} = b^3a2a6=a4,b1b4=b3
So the result is a4b3a^4 b^3a4b3.
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