12 multiple-choice questions, progressively harder.
Solve 2x=322^x = 322x=32.
Solution
Correct answer: C
Write 323232 as a power of 222 and match the exponents.
2x=32=25 ⇒ x=52^x = 32 = 2^5 \;\Rightarrow\; x = 52x=32=25⇒x=5
For large xxx, which grows larger: the exponential 2x2^x2x or the power x100x^{100}x100?
Correct answer: B
An exponential with base greater than 111 eventually outgrows every power function, no matter how large the fixed exponent.
2x overtakes x100 for large enough x2^x \;\text{overtakes}\; x^{100} \;\text{for large enough } x2xovertakesx100for large enough x
The power leads at first, but the constant ratio of the exponential wins in the long run.
Solve (12)x=8\left(\tfrac{1}{2}\right)^x = 8(21)x=8.
Write both sides as powers of 222. Since 12=2−1\tfrac{1}{2} = 2^{-1}21=2−1 and 8=238 = 2^38=23:
(2−1)x=23 ⇒ 2−x=23 ⇒ −x=3 ⇒ x=−3\left(2^{-1}\right)^x = 2^3 \;\Rightarrow\; 2^{-x} = 2^3 \;\Rightarrow\; -x = 3 \;\Rightarrow\; x = -3(2−1)x=23⇒2−x=23⇒−x=3⇒x=−3
A culture triples every 222 hours, starting at 101010. What is the count after 666 hours?
Correct answer: A
Six hours is three tripling periods, so multiply by 33=273^3 = 2733=27.
10⋅33=10⋅27=27010 \cdot 3^3 = 10 \cdot 27 = 27010⋅33=10⋅27=270
One list is exponential. List A is 1,4,7,101, 4, 7, 101,4,7,10; list B is 1,4,16,641, 4, 16, 641,4,16,64. Which is exponential?
Correct answer: D
List A adds 333 each step, a constant difference, so it is linear. List B multiplies by 444 each step.
41=164=6416=4\frac{4}{1} = \frac{16}{4} = \frac{64}{16} = 414=416=1664=4
A constant ratio makes list B the exponential one.
Given 2m=10242^{m} = 10242m=1024 (so m=10m = 10m=10), find 2m+32^{m+3}2m+3.
Adding 333 to the exponent multiplies by 23=82^3 = 823=8.
2m+3=2m⋅23=1024⋅8=81922^{m+3} = 2^m \cdot 2^3 = 1024 \cdot 8 = 81922m+3=2m⋅23=1024⋅8=8192
Every exponential function is one-to-one, which guarantees it has what?
A one-to-one function never repeats an output, so every output comes from exactly one input and the pairing can be reversed.
one-to-one ⇒ an inverse function exists\text{one-to-one} \;\Rightarrow\; \text{an inverse function exists}one-to-one⇒an inverse function exists
This inverse function is named and studied later in the chapter.
Which function decreases and passes through (0,1)(0, 1)(0,1)?
Decreasing needs a base between 000 and 111, and passing through (0,1)(0, 1)(0,1) needs no leading multiplier other than 111.
f(0)=(0.4)0=1and0<0.4<1f(0) = (0.4)^0 = 1 \quad\text{and}\quad 0 < 0.4 < 1f(0)=(0.4)0=1and0<0.4<1
The function 4⋅(0.5)x4 \cdot (0.5)^x4⋅(0.5)x also decays but has f(0)=4f(0) = 4f(0)=4, and 4x4^x4x grows.
For f(x)=2xf(x) = 2^xf(x)=2x, simplify f(x+1)−f(x)f(x+1) - f(x)f(x+1)−f(x).
Use 2x+1=2⋅2x2^{x+1} = 2 \cdot 2^x2x+1=2⋅2x, then factor.
2x+1−2x=2⋅2x−2x=2x(2−1)=2x2^{x+1} - 2^x = 2 \cdot 2^x - 2^x = 2^x(2 - 1) = 2^x2x+1−2x=2⋅2x−2x=2x(2−1)=2x
For f(x)=bxf(x) = b^xf(x)=bx with b>1b > 1b>1, what does the graph approach as x→−∞x \to -\inftyx→−∞?
Large negative exponents give tiny positive reciprocals that shrink toward 000.
bx→0 as x→−∞b^{x} \to 0 \;\text{as}\; x \to -\inftybx→0asx→−∞
The curve flattens toward the horizontal asymptote y=0y = 0y=0 but never reaches it.
Solve 4x=84^x = 84x=8 by writing both sides as powers of 222.
Since 4=224 = 2^24=22 and 8=238 = 2^38=23:
(22)x=23 ⇒ 22x=23 ⇒ 2x=3 ⇒ x=32\left(2^2\right)^x = 2^3 \;\Rightarrow\; 2^{2x} = 2^3 \;\Rightarrow\; 2x = 3 \;\Rightarrow\; x = \tfrac{3}{2}(22)x=23⇒22x=23⇒2x=3⇒x=23
A quantity halves each step from 484848: 48,24,12,…48, 24, 12, \ldots48,24,12,…. What is the first value below 555?
Continue halving until a term drops under 555.
48→24→12→6→348 \to 24 \to 12 \to 6 \to 348→24→12→6→3
The value 666 is still above 555; the next value, 333, is the first below 555.
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