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2.2 Exponents and Logarithms

Q1

Topic
2.2 Exponents and Logarithms
Tag
Exponents; Logarithms; Graph; Asymptotes; Domain; Range; Functions; Exponential rules; Logarithm rules
Source
N17/5/MATHL/HP1/ENG/TZ0/XX/1
Question Text
Solve the equation log⁡2(x+3)+log⁡2(x−3)=4 \log _2(x+3)+\log _2(x-3)=4
Total Mark
5
Correct Answer
5
Explanation
n/a
Mark Scheme
Step 1: Simplify by applying logarithmic properties In this case two logarithms are being added, log⁡2((x+3)×(x−3))=log⁡2(x2−9)=4\log _2((x+3) \times(x-3))=\log _2\left(x^2-9\right)=4 The equation can be further simplified by converting into its exponential form. Notice it becomes a quadratic. x2−9=24x^2-9=2^4, rearranging gives, x2=25x^2=25 As x+3x+3, x−3>0x-3>0, x=5x=5

Q2

Topic
2.2 Exponents and Logarithms
Tag
Exponents; Logarithms; Graph; Asymptotes; Domain; Range; Functions; Exponential rules; Logarithm rules
Source
M17/5/MATHL/HP1/ENG/TZ1/XX/1
Question Text
Find the solution of log⁡2x−log⁡27=3+log⁡23\log _2 \mathrm{x}-\log _2 7=3+\log _2 3
Total Mark
4
Correct Answer
168
Explanation
n/a
Mark Scheme
Rearrange the equation to collect all log terms, log⁡2x−log⁡27−log⁡23=3log⁡2x21=3\log _2 x-\log _2 7-\log _2 3=3 \\ \log _2 \frac{x}{21}=3 Convert to exponential form, x21=23x=168\frac{x}{21}=2^3 \\ x=168

Q3

Topic
2.2 Exponents and Logarithms
Tag
Exponents; Logarithms; Graph; Asymptotes; Domain; Range; Functions; Exponential rules; Logarithm rules
Source
N16-TZ0-P1-7(HL)
Question Text
The equation 2x+2−4x=32^{x+2}-4^x=3 has two solutions. Given that one solution can be written aslog⁡ab\log _a b where aa and bb are prime numbers (i) compute the value of aa Total Mark: 3 Correct Answer: 2 Explanation: n/a Mark Scheme: Use the substitution 2x=A2^x=A 4A−A2=3A2−4A−3=0(A−3)(A−1)=0A=1 or 34 A-A^2=3 \\ A^2-4 A-3=0 \\ (A-3)(A-1)=0 \\ A=1 \text { or } 3 So, 2x=1 or 3x=0 or log⁡232^x=1 \text { or } 3 \\ x=0 \text { or } \log _2 3 (ii) compute the value of bb Total Mark: 2 Correct Answer: 3 Explanation: n/a Mark Scheme: Use the substitution 2x=A2^x=A 4A−A2=3A2−4A−3=0(A−3)(A−1)=0A=1 or 34 A-A^2=3 \\ A^2-4 A-3=0 \\ (A-3)(A-1)=0 \\ A=1 \text { or } 3 So, 2x=1 or 3x=0 or log⁡232^x=1 \text { or } 3 \\ x=0 \text { or } \log _2 3

Q4

Topic
2.2 Exponents and Logarithms
Tag
Exponents; Logarithms; Graph; Asymptotes; Domain; Range; Functions; Exponential rules; Logarithm rules
Source
M16-TZ1-P1-6(HL)
Question Text
Integer values mm and nn satisfy m+nlog34=10log⁡912m+n \mathrm{ log}_3 4=10 \log _9 12. (i) compute the value of mm Total Mark: 3 Correct Answer: 5 Explanation: n/a Mark Scheme: Rearrange to collect the log terms, m=10log⁡912−nlog⁡34m=5log⁡312−nlog⁡34m=log⁡31254−nm=10 \log _9 12-n \log _3 4 \\ m=5 \log _3 12-n \log _3 4 \\ m=\log _3 12^5 4^{-n} Rewrite in exponential form, 3m=1254−n3m4n=35453^m=12^5 4^{-n} \\ 3^m 4^n=3^5 4^5 (ii) compute the value of nn Total Mark: 2 Correct Answer: 5 Explanation: n/a Mark Scheme: Rearrange to collect the log terms, m=10log⁡912−nlog⁡34m=5log⁡312−nlog⁡34m=log⁡31254−nm=10 \log _9 12-n \log _3 4 \\ m=5 \log _3 12-n \log _3 4 \\ m=\log _3 12^5 4^{-n} Rewrite in exponential form, 3m=1254−n3m4n=35453^m=12^5 4^{-n} \\ 3^m 4^n=3^5 4^5

Q5

Topic
2.2 Exponents and Logarithms
Tag
Exponents; Logarithms; Graph; Asymptotes; Domain; Range; Functions; Exponential rules; Logarithm rules
Source
M15-TZ2-P1-9(HL)
Question Text
Given that log⁡2x=4log⁡x2 \log _2 x=4 \log _x 2, find the integer solution.
Total Mark
4
Correct Answer
4
Explanation
n/a
Mark Scheme
Using change of base, log⁡x2=log⁡22log⁡2x\log _x 2=\frac{\log _2 2}{\log _2 x} So, log⁡2x=4log⁡2x(log⁡2x)2=4 \log _2 x=\frac{4}{\log _2 x} \\ \left(\log _2 x\right)^2=4 To obtain the integer solution, log⁡2x=2x=4 \begin{gathered} \log _2 x=2 \\ x=4 \end{gathered}

Q6

Topic
2.2 Exponents and Logarithms
Tag
Exponents; Logarithms; Graph; Asymptotes; Domain; Range; Functions; Exponential rules; Logarithm rules
Source
M14/5/MATHL/HP1/ENG/TZ1/X/3
Question Text
Consider k=log⁡24×log⁡46×log⁡68×…×log⁡332\mathrm{k}=\log _2 4 \times \log _4 6 \times \log _6 8 \times \ldots \times \log _3 32. Given that k∈Zk \in Z, find the value of kk.
Total Mark
4
Correct Answer
4
Explanation
n/a
Mark Scheme
Using change of base, k=log⁡24log⁡22×log⁡26log⁡24×log⁡28log⁡26×…×log⁡232log⁡230 k=\frac{\log _2 4}{\log _2 2} \times \frac{\log _2 6}{\log _2 4} \times \frac{\log _2 8}{\log _2 6} \times \ldots \times \frac{\log _2 32}{\log _2 30} Notice that the values cancel out, k=log⁡232log⁡22=log⁡232=5k=\frac{\log _2 32}{\log _2 2}=\log _2 32=5

Q7

Topic
2.2 Exponents and Logarithms
Tag
Exponents; Logarithms; Graph; Asymptotes; Domain; Range; Functions; Exponential rules; Logarithm rules
Source
M14/5/MATHL/HP1/ENG/TZ2/XX/2
Question Text
The solution to the equation 2x−1=63x2^{x-1}=6^{3 x} can be written as ln⁡aln⁡b\frac{\ln a}{\ln b} where a,b∈Z+a, b \in \mathbb{Z}^{+}. Find the value of a+ba+b.
Total Mark
5
Correct Answer
110
Explanation
n/a
Mark Scheme
2x+1=(2×3)3x2x+1=23x×33x2−2x+1=33xln⁡(2−2x+1)=ln⁡(33x)(−2x+1)ln⁡2=(3x)ln⁡3x(3ln⁡3+2ln⁡2)=ln⁡2x=ln⁡2ln⁡1082^{x+1}=(2 \times 3)^{3 x} \\ 2^{x+1}=2^{3 x} \times 3^{3 x} \\ 2^{-2 x+1}=3^{3 x} \\ \ln \left(2^{-2 x+1}\right)=\ln \left(3^{3 x}\right) \\ (-2 x+1) \ln 2=(3 x) \ln 3 \\ x(3 \ln 3+2 \ln 2)=\ln 2 \\ x=\frac{\ln 2}{\ln 108}

Q8

Topic
2.2 Exponents and Logarithms
Tag
Exponents; Logarithms; Graph; Asymptotes; Domain; Range; Functions; Exponential rules; Logarithm rules
Source
N13-TZ0-P1-9(HL)
Question Text
One solution to the equation xlog⁡x=10(log⁡x)3x^{\log x}=10^{(\log x)^3} is 11 , find another solution.
Total Mark
5
Correct Answer
10
Explanation
n/a
Mark Scheme
xlog⁡x=(10log⁡x)(log⁡x)2xlog⁡x=x(log⁡x)2log⁡x=(log⁡x)2log⁡x=1x=10x^{\log x}=\left(10^{\log x}\right)^{(\log x)^2} \\ x^{\log x}=x^{(\log x)^2} \\ \log x=(\log x)^2 \\ \log x=1 \\ x=10

Q9

Topic
2.2 Exponents and Logarithms
Tag
Exponents; Logarithms; Graph; Asymptotes; Domain; Range; Functions; Exponential rules; Logarithm rules
Source
M19/5/MATHL/HP1/ENG/TZ2/XX/7
Question Text
2+log⁡22x=log⁡23+2log⁡2y1+log⁡5x=log⁡5(7y+1) 2+\log _2 2 \mathrm{x}=\log _2 3+2 \log _2 \mathrm{y} \\ 1+\log _5 \mathrm{x}=\log _5(7 \mathrm{y}+1) (a) compute the value of xx Total Mark: 4 Correct Answer: 3 Explanation: n/a Mark Scheme: Simplify each equation, log⁡24x=log⁡23y24x=3y2log⁡55x=log⁡5(7y+1)5x=7y+1 \log _2 4 x=\log _2 3 y^2 \\ 4 x=3 y^2 \\ \log _5 5 x=\log _5(7 y+1) \\ 5 x=7 y+1 Replacing xx, 154y2=7y+115y2−28y−4=0(y−2)(15y+2=0)y=2 \frac{15}{4} y^2=7 y+1 \\ 15 y^2-28 y-4=0 \\ (y-2)(15 y+2=0) \\ y=2 As 5x=7y+15 x=7 y+1 x=3 x=3 (b) compute the value of yy Total Mark: 3 Correct Answer: 2 Explanation: n/a Mark Scheme: Simplify each equation, log⁡24x=log⁡23y24x=3y2log⁡55x=log⁡5(7y+1)5x=7y+1 \log _2 4 x=\log _2 3 y^2 \\ 4 x=3 y^2 \\ \log _5 5 x=\log _5(7 y+1) \\ 5 x=7 y+1 Replacing xx, 154y2=7y+115y2−28y−4=0(y−2)(15y+2=0)y=2 \frac{15}{4} y^2=7 y+1 \\ 15 y^2-28 y-4=0 \\ (y-2)(15 y+2=0) \\ y=2

Q10

Topic
2.2 Exponents and Logarithms
Tag
Exponents; Logarithms; Graph; Asymptotes; Domain; Range; Functions; Exponential rules; Logarithm rules
Source
M18/5/MATHL/HP1/ENG/TZ1/XX/5
Question Text
Find the integer solution of the equation (ln⁡x)2−(ln⁡3)(ln⁡x)=2(ln⁡3)2(\ln x)^2-(\ln 3)(\ln x)=2(\ln 3)^2 .
Total Mark
6
Correct Answer
9
Explanation
n/a
Mark Scheme
(ln⁡x)2−(ln⁡3)(ln⁡x)−2(ln⁡3)2=0(ln⁡x−2ln⁡3)(ln⁡x+ln⁡3)=0ln⁡x=ln⁡9x=9\begin{aligned} & (\ln x)^2-(\ln 3)(\ln x)-2(\ln 3)^2=0 \\ & (\ln x-2 \ln 3)(\ln x+\ln 3)=0 \\ & \ln x=\ln 9 \\ & x=9\end{aligned}

Q11

Topic
2.2 Exponents and Logarithms
Tag
Exponents; Logarithms; Graph; Asymptotes; Domain; Range; Functions; Exponential rules; Logarithm rules
Source
M18/5/MATHL/HP1/ENG/TZ2/XX/11
Question Text
Find the solution to the equation log⁡2x+log⁡4x+log⁡8x+log⁡64x=6\log _2 x+\log _4 x+\log _8 x+ \log _{64}x =6
Total Mark
5
Correct Answer
8
Explanation
n/a
Mark Scheme
Using change of base, log⁡2x+log⁡4x+log⁡8x+log⁡64x=log⁡2x+log⁡2xlog⁡24+log⁡2xlog⁡28+log⁡2xlog⁡264\log _2 x+\log _4 x+\log _8 x+\log _{64} x=\log _2 x+\frac{\log _2 x}{\log _2 4}+\frac{\log _2 x}{\log _2 8}+\frac{\log _2 x}{\log _2 64} Simplify, log⁡2x+log⁡2x2+log⁡2x3+log⁡2x6=2log⁡2x=6log⁡2x=3x=8\log _2 x+\frac{\log _2 x}{2}+\frac{\log _2 x}{3}+\frac{\log _2 x}{6}=2 \log _2 x=6 \\ \log _2 x=3 \\ x=8