Logarithms : Exercise (3.3) Solutions
1. Replace ◻ with the appropriate number.
(a) log324=log36+log3◻(b) log524=log560+log5◻(c) log2◻=3log23(d) log109=◻log103(e) log85=log8◻−log811
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(a) 4(b) 25(c) 27(d) 2(e) 55
2. Write each expression as a single logarithm.
(a) logb20+logb57−logb241(b) 3logb8−12logb12(c) logbx−2logby−logba(d) log23+log415
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(a) logb20+logb57−logb241 = logb20×57241 = logb1140241(b) 3logb8−12logb12 = logb83−logb1212 = logb8×8×8√12 = logb8×8×82√3 = logb256√3 = logb256√33(c) logbx−2logby−logba = logbx−logby2−logba = logbxay2(d) log23+log415 Let log415=x, then 15=4x 15=22x ∴ 2x=log215 ∴ x=12log215 ∴ log415=log2√15 ∴ log23+log415 = log23+log2√15 = log23√15
3. Write each expression in terms of logb2,logb3 and logb5.
(a) logb8(b) logb15(c) logb270(d) logb273√516(e) logb2163√32(f) logb(648√125)
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(a)logb8=logb23=3logb2(b)logb15=logb(3×5)=logb3+logb5(c)logb270=logb(2×33×5) =logb2+logb33+logb5 =logb2+3logb3+logb5(d)logb273√516=logb33×51324 =logb33+logb513−logb24 =3logb3+13logb5−4logb2(e)logb2163√32=logb23×33253 =logb(243×33) =43logb2+3logb3(f)logb(648√125)=logb(23×34×532) =logb23+logb34+logb532 =3logb2+4logb3+32logb5
4. Evaluate each expression.
(a) log2128(b) log3814(c) log128(d) log82(e) log3√381(f) log3√3log381(g) log225log25(h) log48
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(a)log2128=log227=7(b)log3814=log3(34)4=log3316=16(c)log128=log1223=log12(12)−3=−3(d)log82=log8813=13(e)log3√381=log331234=log33−72=−72(f)log3√3log381=log3312log334=124=18(g)log225log25=log252log25=2log25log25=2(h)log48=log4√64=log4432=32
5. Use log102=0.3010 and log103=0.4771 to evaluate each of the following expressions.
(a) log106(b) log101.5(c) log10√3(d) log104(e) log104.5(f) log108(g) log1018(h) log105
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log102=0.3010, log103=0.4771(a)log106=log10(2×3) =log102+log103 =0.3010+0.4771 =0.7781(b)log101.5=log1032 =log103−log102 =0.4771−0.3010 =0.1761(c)log10√3 =log10312 =12log103 =12×0.4771 =0.2386(d)log104=log1022 =2log102 =2 (0.3010) =0.6020(e)log104.5=log1092 =log10322 =2log103−log102 =2(0.4771)−0.3010 =0.6532(f)log108 =log1023 =3log102 =3(0.3010) =0.9030(g)log1018=log10(2×32) =log102+2log103 =0.3010+2(0.4771)(h)log105=log10102 =log1010−log102 =1−0.3010 =0.6990
6. Solve the following equations for x.
(a) loga185+loga103−loga67=logax(b) logbx=2−a+logb(a2bab2)(c) logx3−logx2=log5x−log4x(d) log10x+log103=log106(e) 8logx=loga32+log2−12loga3−log2a4
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(a)loga185+loga103−loga67=logax loga(185×10367)=logax x=185×103×76 x=14(b)logbx=2−a+logb(a2bab2) logbx=2−a+logba2+logbba−logbb2 logbx=2−a+logba2+a−2 logbx=logba2 x=a2(c)logx3−logx2=log5x−log4x logx3x2=log5x4x logx=log54 x=54(d)log10x+log103=log106 log103x=log106 3x=6 x=2(e)8logx=loga32+log2−12loga3−log2a4 logx8=loga32+log2−loga32−(log2−loga4) logx8=loga4 x8=a4 x=√a
7. Given that log105=0.6990 and log10x=0.2330. What is the value of x?
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log105=0.6990 log10x=0.2330 ∴ 3log10x=0.6990 ∴ 3log10x=log105 ∴ log10x=13log105 ∴ log10x=log10513 ∴ x=513=3√5
8. Show that if logeI=−RLt+logeI0 then I=I0e−RtL
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logeI=−RLt+logeI0logeI−logeI0=−RtLlogeII0=−RtLII0=e−RtL∴ I=I0e−RtL
9. Show that if logby=12logbx+c then y=bc√x
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logby=12logbx+clogby−12logbx=clogby−logbx12=clogby√x=cy√x=bcy= bc√x
10. Show that
(a) 14log108+14log102=log102(b) 4log103−2log103+1=log1090
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(a) 14log108+14log102 =14(log108+log102) =14log10(8×2) =14log1016 =14log1024 =14×4log102 =log102(b) 4log103−2log103+1 =2log103+log1010 =log1032+log1010 =log10(32×10) =log1090
11. Show that
(a) a2loga3+b3logb2=17(b) 3log61296=2log44096
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(a) a2loga3+b3logb2 =aloga32+blogb23 =32+23 =17(b) 3log61296=3log664 =3×4log66 =12 2log44096=2log446 =2×6log44 =12 ∴ 3log61296=2log44096
12. Given that log1012=1.0792 and log1024=1.3802, deduce the values of log102 and log106.
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log1012=1.0792log1024=1.3802log102=log102412 =log1024−log1012 =1.3802−1.0792 =0.3010log106=log10122 =log1012−log102 =1.0792−0.3010 =0.7782
13. If logxa=5 and logx3a=9, find the values of a and x.
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logxa=5∴ a=x5logx3a=9∴ 3a=x9∴ 3x5=x9∴ x4=3∴ x=314=4√3∴ a=(314)5=354=34√3
14. (a) If log102=a, find log108+log1025 in terms of a.
(b) If a=10x and b=10y, express log10(a4b3) in terms of x and y.
(b) If a=10x and b=10y, express log10(a4b3) in terms of x and y.
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(a) log102=a (given) log108+log1025=log108+log101004=log108+log10100−log104=log1023+log10102−log1022=3log102+2log1010−2log102=log102+2=a+2(b) a=10xb=10y}(given) log10(a4b3)=log10((10x)4(10y)3) =log10((104x)(103y)) =log10104x+3y =4x+3y
15. (a) If log2(4x−4)=2, find the value of log4x.
(b) Prove that if 12log3M+3log3N=1 then MN6=9.
(b) Prove that if 12log3M+3log3N=1 then MN6=9.
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(a) log2(4x−4)=2 4x−4=22 4x=8 x=2 x=412 log4x=12(b) 12log3M+3log3N=1 log3M12+log3N3=1 log3(M12N3)=1 M12N3=3 Squaring both sides. MN6=9
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