Solve the system \(\begin{cases}3y+2^x=-13,\\2^x-y=15.\end{cases}\) If \((x_0,y_0)\) is its solution, find \(x_0+y_0\).
Solution
From the second equation, \(2^x=15+y\). Substituting into the first gives \(3y+15+y=-13\), so \(y=-7\). Then \(2^x=8\), hence \(x=3\). Therefore \(x_0+y_0=-4\). Answer: A.
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2
Task
One correct answer
Which interval contains the solution of \(3^x=\dfrac1{27}\)?
Solution
\(\dfrac1{27}=3^{-3}\). Thus \(x=-3\), and \(-3\in(-5,-2]\). Answer: B.
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3
Task
One correct answer
Which interval contains the solution of \(3^x=30\)?
Solution
Since \(3^3=27\lt30\lt81=3^4\) and \(3^x\) is increasing, \(3\lt x\lt4\). Answer: C.
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4
Task
Numeric answer
Solve \(3^x\cdot4^x=(12^{x+1})^5\).
Solution
\(3^x\cdot4^x=12^x\), while \((12^{x+1})^5=12^{5x+5}\). Thus \(x=5x+5\), giving \(x=-\dfrac54=-1.25\).
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5
Task
Numeric answer
Solve \(\begin{cases}3^{x-2y}=\dfrac13,\\3^x+3^{2y}=4\sqrt3.\end{cases}\) For the solution \((x_0,y_0)\), find \(x_0y_0\).
Solution
The first equation gives \(x-2y=-1\), so \(2y=x+1\). Hence \(3^{2y}=3^{x+1}=3\cdot3^x\). The second equation becomes \(4\cdot3^x=4\sqrt3\), so \(x=\dfrac12\). Then \(y=\dfrac34\), and \(x_0y_0=\dfrac38=0.375\).
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6
Task
One correct answer
Solve \(\begin{cases}x+y=5,\\4^x=16^{-1}.\end{cases}\) If \((x_0,y_0)\) is the solution, find \(x_0y_0\).
Solution
\(16^{-1}=4^{-2}\), so \(x=-2\). From \(x+y=5\), \(y=7\). Thus \(x_0y_0=-14\). Answer: B.
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7
Task
One correct answer
Which interval contains the solution of \(5^{x+1}=125\)?
Solution
\(125=5^3\), so \(x+1=3\) and \(x=2\). Therefore \(2\in[0,3)\). Answer: A.
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8
Task
One correct answer
Which interval contains the solution of \(3^{x+4}=27\)?
Solution
\(27=3^3\), so \(x+4=3\) and \(x=-1\). Therefore \(-1\in[-2,0)\). Answer: B.
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9
Task
One correct answer
Find the sum of the roots of \(\left(3^{2x^2-29}-27\right)\sqrt[4]{5x+18}=0\).
Solution
Domain: \(x\ge-3.6\). The first factor is zero when \(3^{2x^2-29}=3^3\), giving \(x=\pm4\); only \(x=4\) is admissible. The fourth-root factor is zero at \(x=-3.6\). Their sum is \(0.4\). Answer: A.
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10
Task
One correct answer
Solve \(4^x=8\).
Solution
\(4=2^2\) and \(8=2^3\). Thus \(2^{2x}=2^3\), so \(x=\dfrac32\). Answer: C.
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11
Task
One correct answer
Which interval contains the solution of \(2^x=\dfrac18\)?
Solution
\(\dfrac18=2^{-3}\), so \(x=-3\). The value \(-3\) lies in \((-4,-2]\). Answer: B.
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12
Task
One correct answer
Which interval contains the solution of \(5^{x+3}=\left(\dfrac1{125}\right)^x\)?
Solution
\(\dfrac1{125}=5^{-3}\), so \(5^{x+3}=5^{-3x}\). Hence \(x+3=-3x\), giving \(x=-\dfrac34=-0.75\). Thus \(x\in(-1,0]\). Answer: C.
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13
Task
One correct answer
Choose the interval containing the solution of \(3^x\cdot4^x=\dfrac1{144}\).
Solution
\(3^x\cdot4^x=12^x\) and \(\dfrac1{144}=12^{-2}\). Hence \(x=-2\), which lies in \([-5,-1)\). Answer: B.
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14
Task
One correct answer
Solve \(\sqrt[3]{8^x}=\sqrt2\cdot\sqrt[3]{2}\).
Solution
\(\sqrt[3]{8^x}=2^x\), while \(\sqrt2\cdot\sqrt[3]{2}=2^{1/2+1/3}=2^{5/6}\). Thus \(x=\dfrac56\). Answer: D.
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15
Task
One correct answer
Solve \(3^x=\dfrac{2\sqrt3}{6}\).
Solution
\(\dfrac{2\sqrt3}{6}=\dfrac{\sqrt3}{3}=3^{-1/2}\). Hence \(x=-\dfrac12=-0.5\). Answer: C.
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16
Task
One correct answer
Solve \(2^{2x}=\dfrac1{2^3}\).
Solution
\(\dfrac1{2^3}=2^{-3}\). Thus \(2x=-3\), so \(x=-\dfrac32=-1.5\). Answer: C.
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17
Task
One correct answer
Which interval contains the solution of \(2^{x+3}-3\cdot2^x=10\sqrt2\)?
Solution
Factor out \(2^x\): \(5\cdot2^x=10\sqrt2\). Thus \(2^x=2\sqrt2=2^{3/2}\), so \(x=1.5\in[1,2)\). Answer: D.
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18
Task
One correct answer
Choose the number that is a root of \(5^{x-2}=25\).
Solution
\(25=5^2\), so \(x-2=2\), giving \(x=4\). Answer: B.
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19
Task
One correct answer
Solve \(3^{7x}=9\). Round the solution to the nearest tenth.
Solution
\(9=3^2\), so \(7x=2\). Thus \(x=\dfrac27\approx0.2857\), which rounds to \(0.3\). Answer: C.
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20
Task
Numeric answer
Solve \(3\cdot4^x+(3x-10)\cdot2^x+3-x=0\). Enter the larger root.
Solution
Let \(t=2^x\gt0\). Then \(4^x=t^2\), so the equation becomes \(3t^2+(3x-10)t+3-x=0\). Grouping terms gives \((3t-1)(t+x-3)=0\). Hence either \(3\cdot2^x=1\), giving \(x=\log_2\dfrac13=-\log_2 3\), or \(2^x+x-3=0\). The function \(2^x+x\) is strictly increasing; \(x=1\) satisfies the equation and is therefore the unique root of this branch. The larger root is \(1\).
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21
Task
One correct answer
Choose the interval containing the solution of \(\left(\dfrac13\right)^{2x-1}=9\).
Solution
Write \(9=3^2=\left(\dfrac13\right)^{-2}\). Then \(2x-1=-2\), so \(x=-\dfrac12\). Hence \(x\in(-1,0]\). Answer: B.
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22
Task
Numeric answer
Solve \(12\cdot5^{x-1}+3\cdot5^x-5^{x+1}=10\).
Solution
Factor out \(5^x\): \(5^x\left(\dfrac{12}{5}+3-5\right)=10\). Thus \(\dfrac25\cdot5^x=10\), so \(5^x=25=5^2\). Hence \(x=2\).
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23
Task
One correct answer
Solve \(0.2^{3x-1}=\sqrt{125}\).
Solution
\(0.2=5^{-1}\) and \(\sqrt{125}=5^{3/2}\). Thus \(-3x+1=\dfrac32\), giving \(x=-\dfrac16\). Answer: D.
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24
Task
One correct answer
Solve \(4^{x-1}-1.5\cdot2^{x+2}+20=0\).
Solution
Let \(t=2^x\gt0\). Then \(4^{x-1}=\dfrac{t^2}{4}\) and \(1.5\cdot2^{x+2}=6t\). We get \(t^2-24t+80=0\), so \(t=4\) or \(20\). Hence \(x=2\) or \(x=\log_2 20\). Answer: C.
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25
Task
One correct answer
Solve \(3\cdot4^x-5\cdot6^x+2\cdot9^x=0\).
Solution
Divide by \(3^{2x}\ne0\): \(3\left(\dfrac23\right)^{2x}-5\left(\dfrac23\right)^x+2=0\). Let \(t=\left(\dfrac23\right)^x\gt0\). Then \(3t^2-5t+2=0\), so \(t=1\) or \(t=\dfrac23\). Hence \(x=0\) or \(x=1\). Answer: E.
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26
Task
One correct answer
Solve \(7^{x+1}=14\left(\dfrac12\right)^{3x+1}\).
Solution
Multiply by \(2^{3x+1}\): \(7^{x+1}2^{3x+1}=14\), i.e. \(14\cdot56^x=14\). Thus \(56^x=1\), so \(x=0\). Answer: C.
Domain: \(x\ne0\). Since \(5^{\frac{x+4}{x}}=5\left(5^{\frac2x}\right)^2\), let \(t=5^{\frac2x}\gt0\). Then \(5t^2-124t-25=0\), giving \(t=-\dfrac15\) or \(25\). Only \(t=25\) is admissible. Hence \(5^{2/x}=5^2\), so \(x=1\). Answer: A.
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28
Task
Numeric answer
Solve \(2^{x-1}=5^{x-1}\).
Solution
Divide by \(5^{x-1}\ne0\): \(\left(\dfrac25\right)^{x-1}=1=\left(\dfrac25\right)^0\). Hence \(x-1=0\), so \(x=1\).
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29
Task
Numeric answer
Find the least value of the parameter \(a\) for which \(2^{\sin^2\left(2\pi x+\frac{5\pi}{4}\right)}=\dfrac{4}{(x-a)^2-6(x-a)+13}\) has a positive root.
Solution
Since \(0\le\sin^2(\cdots)\le1\), the left side lies in \([1,2]\). Also \((x-a)^2-6(x-a)+13=(x-a-3)^2+4\ge4\), so the right side lies in \((0,1]\). Equality is possible only when both sides equal \(1\). Thus \(\sin^2\left(2\pi x+\frac{5\pi}{4}\right)=0\) and \(x=a+3\). The first condition gives \(x=\dfrac n2-\dfrac58\), \(n\in\mathbb Z\). For a positive root, \(n\gt\dfrac54\); the smallest integer is \(n=2\). Hence \(x=\dfrac38\) and \(a=x-3=-\dfrac{21}{8}=-2.625\).
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30
Task
Extended response
Consider the system \[\begin{cases}ax^2+ax+3^{2+y^2}=27,\\x+3^{1+y^2}=8,\end{cases}\] where \(x,y\) are variables and \(a\) is a parameter. **1.** Solve the system for \(a=0\). **2.** Determine all solutions depending on \(a\).
Solution
**1. Case \(a=0\).** The first equation gives \(3^{2+y^2}=27=3^3\), hence \(y^2=1\) and \(y=\pm1\). The second equation then gives \(x+9=8\), so \(x=-1\). The solutions are \((-1,1)\) and \((-1,-1)\).
**2. Arbitrary \(a\).** From the second equation, \(3^{1+y^2}=8-x\), hence \(3^{2+y^2}=3(8-x)\). Substitution into the first equation gives \(ax^2+x(a-3)-3=0\), whose discriminant is \(D=(a+3)^2\). For \(a\ne0\), the roots are \(x_1=\dfrac3a\) and \(x_2=-1\). For \(x=-1\), we always obtain \(y=\pm1\). For \(x=\dfrac3a\), \(3^{y^2}=\dfrac{8a-3}{3a}\), so \(y^2=\log_3\dfrac{8a-3}{3a}\). Real \(y\) exist exactly when \(\dfrac{8a-3}{3a}\ge1\), equivalently \(\dfrac{5a-3}{3a}\ge0\), which gives \(a\in(-\infty,0)\cup[0.6,+\infty)\). Thus for \(a\in[0,0.6)\) the only solutions are \((-1,\pm1)\); outside that interval we also have \(\left(\dfrac3a,\pm\sqrt{\log_3\dfrac{8a-3}{3a}}\right)\).
