By the definition of logarithm, \(\log_3x=-1\) is equivalent to \(x=3^{-1}\). Hence \(x=\dfrac13\). Answer: A.
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2
Task
Numeric answer
Solve \(\log_6(x-3)+\log_6(x-8)=2\). If there is one root, enter it; if there are two roots, enter their sum.
Solution
Domain: \(x\gt8\). Combine the logarithms: \(\log_6((x-3)(x-8))=2\), hence \((x-3)(x-8)=36\). This gives \(x^2-11x-12=0\), with roots \(12\) and \(-1\). Only \(12\) belongs to the domain.
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3
Task
Numeric answer
Solve \(|3\lg x+1|-|\lg x-3|=2\). If there is one root, enter it; if there is more than one, enter the sum of all roots.
Solution
Domain: \(x\gt0\). Let \(t=\lg x\). Then \(|3t+1|-|t-3|=2\), with critical points \(-\frac13\) and \(3\). For \(t\lt-\frac13\), we get \(t=-3\), so \(x=0.001\). For \(-\frac13\le t\lt3\), we get \(t=1\), so \(x=10\). There is no solution in the third interval. The sum is \(10.001\).
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4
Task
Numeric answer
Solve \((\log_5x)^2+\log_5x=2\). If there is one root, enter it; if there are several, enter their sum; if there are none, enter \(100\).
Solution
Domain: \(x\gt0\). Let \(t=\log_5x\). Then \(t^2+t-2=0\), so \(t=-2\) or \(t=1\). Hence \(x_1=0.04\) and \(x_2=5\). Their sum is \(5.04\).
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5
Task
One correct answer
Which interval contains the solution of \(\log_2x=2\log_2 3\)?
Solution
\(2\log_2 3=\log_2 9\), so \(x=9\). Hence \(9\in(8,10]\). Answer: E.
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6
Task
One correct answer
Choose the number that is a root of \(-\log_2x=3\).
Solution
\(-\log_2x=3\Rightarrow\log_2x=-3\). Thus \(x=2^{-3}=\dfrac18\). Answer: D.
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7
Task
One correct answer
Choose the interval containing the solution of \(\log_{\frac13}(x+1)=-2\).
Solution
\(x+1=(1/3)^{-2}=9\), so \(x=8\). Thus \(8\in(7,9]\). Answer: E.
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8
Task
Numeric answer
Solve \(\log_4x\cdot\left(\log_4x+\log_4\dfrac1{16}\right)=3\). If there is one root, enter it; if there are several, enter their sum; if there are none, enter \(100\).
Solution
Domain: \(x\gt0\). We have \(\log_4(1/16)=-2\). Let \(t=\log_4x\). Then \(t(t-2)=3\), so \(t=-1\) or \(t=3\). Hence \(x_1=0.25\), \(x_2=64\), and their sum is \(64.25\).
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9
Task
Numeric answer
Solve \(\log_{0.4}(5x^2-8)=\log_{0.4}(-3x)\). If there is one root, enter it; if there are several, enter their sum.
Solution
Domain: \(5x^2-8\gt0\) and \(-3x\gt0\), so in particular \(x\lt0\). Equating the arguments gives \(5x^2-8=-3x\), with roots \(1\) and \(-1.6\). Only \(-1.6\) belongs to the domain.
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10
Task
Numeric answer
Solve \(\log_{x^2-1}(x^3+6)=\log_{x^2-1}(4x^2-x)\). Enter the sum of the roots.
Solution
The logarithm requires the base \(x^2-1\) to be positive and different from \(1\), and both arguments to be positive. On the domain, equality of the logarithms gives \(x^3+6=4x^2-x\), i.e. \((x+1)(x-2)(x-3)=0\). The value \(x=-1\) is inadmissible, while \(x=2\) and \(x=3\) are valid. Their sum is \(5\).
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11
Task
One correct answer
Which interval contains the solution of \(\log_3 x=2\)?
Solution
By the definition of logarithm, \(x=3^2=9\). The number \(9\) lies in \((8,11]\). Answer: E.
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12
Task
Numeric answer
Solve \(\log_2x+\log_2(x-7)=3\). If there is one root, enter it; if there are several, enter their sum.
Solution
Domain: \(x\gt7\). Combining the logarithms gives \(\log_2(x(x-7))=3\), hence \(x^2-7x-8=0\). The candidates are \(8\) and \(-1\); only \(8\) is in the domain.
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13
Task
One correct answer
Solve \(\log_x(2x^2-3x-4)=2\).
Solution
The logarithm base requires \(x\gt0\) and \(x\ne1\). By definition, \(2x^2-3x-4=x^2\), so \((x-4)(x+1)=0\). The candidates are \(4\) and \(-1\); only \(4\) is an admissible base. Answer: B.
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14
Task
Numeric answer
Solve \(\begin{cases}2^{2y-x}=32,\\ \log_{\frac12}(y-x)=-2.\end{cases}\). Enter the product \(x_0y_0\) for the solution \((x_0,y_0)\).
Solution
The first equation gives \(2y-x=5\). The second gives \(y-x=4\). Hence \(y=1\) and \(x=-3\). Therefore \(x_0y_0=-3\).
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15
Task
One correct answer
Solve \(x-1+\log_4 3=\log_4(5^x-4^{x-1})\).
Solution
Write the left side as a logarithm: \(x-1+\log_4 3=\log_4(3\cdot4^{x-1})\). Thus \(3\cdot4^{x-1}=5^x-4^{x-1}\), so \(4^x=5^x\). Hence \((4/5)^x=1\), giving \(x=0\). The logarithm argument is positive at this value. Answer: C.
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16
Task
One correct answer
Which of the following numbers is a root of \(\log_4(x-1)=3\)?
Solution
By definition, \(x-1=4^3=64\). Hence \(x=65\). Answer: D.
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17
Task
One correct answer
Choose the interval containing the solution of \(\log_{64}x=\dfrac12\).
Solution
By definition, \(x=64^{1/2}=8\). Hence \(8\in(6,32)\). Answer: D.
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18
Task
One correct answer
Solve \(4+\log_{\frac12}x=0\).
Solution
\(\log_{1/2}x=-4\), so \(x=(1/2)^{-4}=16\). Answer: E.
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19
Task
One correct answer
Solve \(\lg(5x-3)=1\).
Solution
We have \(5x-3=10\). Thus \(5x=13\), so \(x=\dfrac{13}{5}=2.6\). Answer: C.
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20
Task
One correct answer
Solve \(\log_2(2x+1)=\log_2(9x+17)-\log_2(x+5)\).
Solution
Domain: \(x\gt-\dfrac12\). Combining the logarithms gives \((2x+1)(x+5)=9x+17\). This simplifies to \(x^2+x-6=0\), so \(x=2\) or \(x=-3\). Only \(x=2\) belongs to the domain. Answer: B.
From \(\log_5(\log_4(\log_3x))=0\), we get \(\log_4(\log_3x)=1\). Hence \(\log_3x=4\), so \(x=3^4=81\). Answer: E.
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22
Task
One correct answer
Solve \((\log_2x)^2-\log_2x^5=4\log_2 64\).
Solution
Domain: \(x\gt0\). Let \(t=\log_2x\). Then \(t^2-5t-24=0\), so \(t=8\) or \(t=-3\). Hence \(x=256\) or \(x=\dfrac18\). Answer: D.
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23
Task
One correct answer
Solve \(\log_8(x-7)+\log_3(7-x)=11\).
Solution
The first logarithm requires \(x\gt7\), while the second requires \(x\lt7\). These conditions are incompatible, so there are no solutions. Answer: A.
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24
Task
One correct answer
Solve \(\log_x7+2\log_x7=3\).
Solution
The base must satisfy \(x\gt0\) and \(x\ne1\). From \(3\log_x7=3\), we get \(\log_x7=1\), hence \(x=7\). Answer: E.
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25
Task
Numeric answer
Find the smallest integer value of the parameter \(a\) for which \(\log_8(x+2)=\log_8(2x-a)\) has a solution.
Solution
Equality of the logarithms gives \(x=a+2\). Both arguments must be positive; after substitution this becomes \(a+4\gt0\), so \(a\gt-4\). The smallest integer value is \(-3\).
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26
Task
Numeric answer
Find all values of the parameter \(a\) for which the product of the roots of \[(\log_2x)^2-(2a^2-a)\log_2x+1-2a=0\] equals \(8\). If there is one such value, enter it; if there are several, enter the smallest.
Solution
Let \(t=\log_2x\). Since \(x_1x_2=2^{t_1+t_2}=8=2^3\), we need \(t_1+t_2=3\). By Vieta, \(t_1+t_2=2a^2-a\), so \(a=-1\) or \(a=1.5\). For \(a=-1\), the quadratic in \(t\) has negative discriminant; for \(a=1.5\), it has real roots. Thus the only admissible value is \(1.5\).
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27
Task
Numeric answer
Find the value of the parameter \(a\) for which a root of \[\lg(\sin5\pi x)=\sqrt{16+a-x}\] lies in \(\left[\dfrac32,2\right]\).
Solution
Since \(0\lt\sin5\pi x\le1\), we have \(\lg(\sin5\pi x)\le0\), while the right-hand side is nonnegative. Thus both sides must be zero. Hence \(\sin5\pi x=1\) and \(16+a-x=0\). In \([3/2,2]\), the first equation gives only \(x=1.7\). Therefore \(a=1.7-16=-14.3\).
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28
Task
Extended response
Consider the equation \[\frac{(x-2)\left(x^2-3(a-1)x+2a^2-3a\right)}{\log_{0.5}(3-2x)+2}=0,\] where \(x\) is the variable and \(a\) is a parameter. 1. Find the admissible values of \(x\). 2. Solve the equation depending on \(a\).
Solution
1. We need \(3-2x\gt0\) and a nonzero denominator. Thus \(x\lt1.5\) and \(x\ne-0.5\), so \(x\in(-\infty,-0.5)\cup(-0.5,1.5)\).
2. The factor \(x-2\) gives no admissible root. The quadratic factor is \((x-a)(x-(2a-3))\). Hence the possible roots are \(x=a\) and \(x=2a-3\), subject to the domain. For \(x=a\), we need \(a\in(-\infty,-0.5)\cup(-0.5,1.5)\). For \(x=2a-3\), we need \(a\in(-\infty,1.25)\cup(1.25,2.25)\). Combining these conditions gives the stated parameter cases.
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29
Task
One correct answer
Choose the interval containing the solution of \(1-\log_{0.5}x=3\).
Solution
From \(1-\log_{0.5}x=3\), we get \(\log_{0.5}x=-2\). Thus \(x=(0.5)^{-2}=4\), which lies in \([3,10)\). Answer: D.
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30
Task
Numeric answer
Find the sum of all integer values of \(a\) for which \(\lg(2ax+5-a)=\lg(4x)\) has no roots.
Solution
The domain requires \(x\gt0\). Equality of the logarithms gives \(x(2a-4)=a-5\). For \(a=2\), this becomes a contradiction, so there is no root. For \(a\ne2\), \(x=\dfrac{a-5}{2a-4}\). The equation has no root when this value is outside the domain, i.e. when \(x\le0\), which gives \(a\in(2,5]\). Including the special case \(a=2\), we get \([2,5]\). The integer values are \(2,3,4,5\), whose sum is \(14\).
