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Logarithms : Exercise (3.3) Solutions



           Properties         For Exponents               For Logarithms  One-to-one Property  If bx=by, then x=y If logbM=logbN, then M=N Product Property bxby=bx+ylogb(MN)=logbM+logbN Quotient Property bxby=bxylogbMN=logbMlogbN Power Property (bx)y=bxylogbNp=plogbN

1.           Replace with the appropriate number.

              (a)  log324=log36+log3(b)  log524=log560+log5(c)  log2=3log23(d)  log109=log103(e)  log85=log8log811

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(a)  4(b)  25(c)  27(d)  2(e)  55

2.           Write each expression as a single logarithm.

              (a)  logb20+logb57logb241(b)  3logb812logb12(c)  logbx2logbylogba(d)  log23+log415

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(a)  logb20+logb57logb241    = logb20×57241    = logb1140241(b)  3logb812logb12    = logb83logb1212    = logb8×8×812    = logb8×8×823    = logb2563    = logb25633(c)  logbx2logbylogba    = logbxlogby2logba    = logbxay2(d)  log23+log415    Let log415=x, then    15=4x    15=22x      2x=log215      x=12log215      log415=log215      log23+log415    =  log23+log215    =  log2315

3.           Write each expression in terms of logb2,logb3 and logb5.

              (a)  logb8(b)  logb15(c)  logb270(d)  logb273516(e)  logb216332(f)  logb(648125)

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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)logb273516=logb33×51324                =logb33+logb513logb24                =3logb3+13logb54logb2(e)logb216332=logb23×33253               =logb(243×33)              =43logb2+3logb3(f)logb(648125)=logb(23×34×532)                    =logb23+logb34+logb532                    =3logb2+4logb3+32logb5

4.           Evaluate each expression.

              (a)  log2128(b)  log3814(c)  log128(d)  log82(e)  log3381(f)  log33log381(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)log3381=log331234=log3372=72(f)log33log381=log3312log334=124=18(g)log225log25=log252log25=2log25log25=2(h)log48=log464=log4432=32

5.           Use log102=0.3010 and log103=0.4771 to evaluate each of the following expressions.

              (a)  log106(b)  log101.5(c)  log103(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              =log103log102              =0.47710.3010              =0.1761(c)log103 =log10312              =12log103              =12×0.4771              =0.2386(d)log104=log1022            =2log102            =2 (0.3010)            =0.6020(e)log104.5=log1092              =log10322              =2log103log102              =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            =log1010log102            =10.3010            =0.6990

6.           Solve the following equations for x.

              (a)  loga185+loga103loga67=logax(b)  logbx=2a+logb(a2bab2)(c)  logx3logx2=log5xlog4x(d)  log10x+log103=log106(e)  8logx=loga32+log212loga3log2a4

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(a)loga185+loga103loga67=logax    loga(185×10367)=logax    x=185×103×76    x=14(b)logbx=2a+logb(a2bab2)    logbx=2a+logba2+logbbalogbb2    logbx=2a+logba2+a2    logbx=logba2    x=a2(c)logx3logx2=log5xlog4x    logx3x2=log5x4x    logx=log54    x=54(d)log10x+log103=log106     log103x=log106     3x=6     x=2(e)8logx=loga32+log212loga3log2a4    logx8=loga32+log2loga32(log2loga4)    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=35

8.           Show that if logeI=RLt+logeI0 then I=I0eRtL

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logeI=RLt+logeI0logeIlogeI0=RtLlogeII0=RtLII0=eRtL I=I0eRtL

9.           Show that if logby=12logbx+c then y=bcx

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logby=12logbx+clogby12logbx=clogbylogbx12=clogbyx=cyx=bcy= bcx

10.           Show that

              (a)  14log108+14log102=log102(b)  4log1032log103+1=log1090

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(a)   14log108+14log102    =14(log108+log102)    =14log10(8×2)    =14log1016    =14log1024    =14×4log102    =log102(b)  4log1032log103+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       =log1024log1012       =1.38021.0792       =0.3010log106=log10122       =log1012log102       =1.07920.3010       =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=43  a=(314)5=354=343

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.

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(a) log102=a (given)   log108+log1025=log108+log101004=log108+log10100log104=log1023+log10102log1022=3log102+2log10102log102=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(4x4)=2, find the value of log4x.

    (b)  Prove that if 12log3M+3log3N=1 then MN6=9.

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(a) log2(4x4)=2   4x4=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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