Tampilkan postingan dengan label Practice Questions 8 Preparation of PAS Questions for Odd Mathematics Subjects Class X (Exponent Functions and Logarithmic Functions). Tampilkan semua postingan
Tampilkan postingan dengan label Practice Questions 8 Preparation of PAS Questions for Odd Mathematics Subjects Class X (Exponent Functions and Logarithmic Functions). Tampilkan semua postingan

Latihan Soal 8 Persiapan PAS Gasal Matematika Peminatan Kelas X (Fungsi Eksponen dan Fungsi Logaritma)

 71.(SPMB '05)Jika3log2=pdan2log7=q,maka14log54=....a.p+3p+qb.2pp+qc.p+3p(q+1)d.p+qp(q+1)e.p(q+1)p+qJawab:c14log54=...log54...log14=...log(2.27)...log(2.7),pilih basis 2mengapa tidak pilih basis selain 2lihat penyebut, di sana terdapat numerus 7pada soal, pasangan numerus 7 adalah 2,makanya basis 2 dipilih, bukan yang lain=2log(2.27)2log(2.7)=2log2+2log272log2+2log7=2log2+2log332log2+2log7=2log2+(3×13log2)2log2+2log7=1+3p1+q=p+3p(q+1).

72.(SPMB '04)Jikaa>1,maka penyelesaian untuk(alog(2x+1))(3loga)=1adalah.... a.1b.2c.3d.4e.5Jawab:d(alog(2x+1))(3loga)=1(3loga)(alog(2x+1))=1(3loga.12)(alog(2x+1))=112(3loga)(alog(2x+1))=13log(2x+1)=22x+1=322x=912x=8x=4.

73.(SPMB '04)Nilai(5log10)2(5log2)25log20adalah.... a.12b.1c.2d.4e.5Jawab:c(5log10)2(5log2)25log20=(5log10+5log2)(5log105log2)5log(20).12=5log(10.2)×5log(102)12×5log20=2×(5log205log20)×5log5=2.1.1=2.

74.(SPMB '03)Jika diketahui bahwa4log4logx4log4log4log16=2maka....a.2logx=8b.2logx=4c.4logx=8d.4logx=16e.16logx=8Jawab:c4log4logx4log4log4log16=24log4logx4log4log4log42=24log4logx4log4log2=24log4logx4log22log21=24log4logx4log(12)=24log4logx22log21=24log4logx(12)=24log4logx+12=24log4logx=212=324logx=4.324logx=(22).324logx=234logx=8.

75.(UMPTN '92)Jikaxmemenuhi persamaan4log4logx4log4log4log16=2maka nilai16logx=....a.4b.2c.1d.2e.4Jawab:a4log4logx4log4log4log16=2menyebabkan4logx=8x=48(lihat pembahasan no.23)maka,16logx=16log48=42log48=82=4.

78.(UMPTN '94)Hasil kali akar-akar persamaan3logx.(2+3logx)=15adalah....a.19b.13c.1d.3e.9Jawab:a3logx.(2+3logx)=15(2+3logx)3logx15=023logx+(3logx)215=0(3logx)2+23logx15=0(3logx1+5)(3logx23)=03logx1+5=0atau3logx13=03logx1=5atau3logx2=3x1=35ataux2=33makax1×x2=35×33=35+3=32=132=19.

79.Diketahui bahwa2log3=pdan3log11=q,maka nilai44log66=....Jawab:44log66=...log66...log44=...log(2×3×11)...log(22×11)=3log2+3log3+3log113log22+3log11=12log3+3log3+3log1123log22+3log11=1p+1+q2p+q=1p+1+q2p+q×pp=1+p+pq2+pq

80.(AIME 1984)Diketahui bahwaxdanyadalah bilangan real yang memenuhi{8logx+4logy2=58logy+4logx2=7Tentukanlah nilai darixyJawab:8logx+4logy2=523logx+22logy2=5....(1)8logy+4logx2=723logy+22logx2=7....(2)selanjutnya,13.2logx+2logy=5|×13|19.2logx+13.2logy=53....(3)2logx+13.2logy=7|×1|2logx+13.2logy=7....(4)saat persamaan(3)(4)=89.2logx=537=163maka2logx=(163)(98)2logx=6x=26x=64Pada persamaan 1 selanjutnya13.2logx+2logy=513.2log26+2logy=513.6+2logy=52+2logy=52logy=52=3y=23=8Jadi,x.y=64.8=512

81.Tentukanlah nilai daria.(22log6)(39log5)(515log2)b.(27log125)(25log164)(64log19)c.(625log19)(7log125)(19log7)Jawab:a.(22log6)(39log5)(515log2)=(22log6)(332log5)(551log2)=(22log6)(33log512)(55log21)=(22log6)(33log5)(55log12)=6×5×12=35b.(27log125)(25log164)(64log19)=(33log53)(52log43)(43log32)=33.(32).(23).3log5.5log4.4log3=1Pembahasan diserahkan kepadaPembaca yang budiman untuk poin c

82.Tentukanlah nilaia+bdimanaadanbadalah bilangan riil positif.7log(1+a2)7log25=7log(2ab15)7log(25+b2)Jawab:7log(1+a2)25=7log(2ab15)(25+b2)diambilpersamaannya, maka(1+a2)25=(2ab15)(25+b2)(1+a2)(25+b2)=25(2ab15){(1+a2) faktor dari25,a>0,aRatau(25+b2) faktor dari 25,b>0,bRjugaNoa(1+a2)b(25+b2)Keterangana+b12510125Memenuhi122750tidak3550tidak

83.Jika60a=3dan60b=5maka hasil dari12(1xy2(1b))Jawab:Perhatikanlah bahwa{60a=360log3=a60b=560log5=bSelanjutnyaUntuk(1ab),maka1ab=160log360log5=60log6060log360log5=60log603×5=60log4=60log22Untuk2(1b),maka2(1b)=2(160log5)=2(60log6060log5)=2(60log605)=2(60log12)=60log122Untuk(1xy2(1b)),maka(1xy2(1b))=60log2260log122=260log2260log12=60log260log12=12log2Jadi,12(1xy2(1b))=1212log2=2

84.Diberikan bilangan riil positifx,y,danzyang memenuhi persamaan2xlog(2y)=22xlog(4z)=24xlog(8yz)0.Jika nilaixy5zdapat dinyatakan dengan12pqdenganpdanqbilangan asli yang saling prima,maka nilai darip+q=....Jawab:2xlog(2y)=22xlog(4z)=24xlog(8yz)0makaxlog(2y)=2xlog(4y)log(2y)×log(2x)=logx×log(4y)...(1)xlog(2y)=4xlog(8yz)log(2y)×log(4x)=logx×log(8yz)....(2)2xlog(4y)=4xlog(8yz)log(4y)×log(4x)=log(2x)×log(8yz)....(3)Perhatikan persamaan(2),yaitu:log(2y)×log(4x)=logx×log(8yz)log(2y)×(log(2x)+log2)=logx×log(8yz)log(2y)×log(2x)+log(2y)×log2=logx×log(8yz)logx×log(4y)+log(2y)×log2=logx×log(8yz)persamaan di atas, persamaan(1)disubstitusikanlog(2y)=logx×log(8yz)logx×log(4y)log2log(2y)=logx×(log8yz4y)log2log(2y)=logx×log(2z)log2......(4)Perhatikan juga persamaan(3),yaitu:log(4y)×log(4x)=log(2x)×log(8yz)(log(2y)+log2)×log(4x)=log(2x)×log(8yz)log(2y)×log(4x)+log2×log(4x)=log(2x)×log(8yz)logx×log(8yz)+log2×log(4x)=log(2x)×log(8yz)di atas, persamaan(2)disubstitusikanlog2×log(4x)=log(2x)×log(8yz)logx×log(8yz)log2×log(4x)=log(8yz)×(log(2x)logx)log2×log(4x)=log(8yz)×(log2xx)log2×log(4x)=log(8yz)×log2log4x=log(8yz)4x=8yzxz=2y....(5)

.dari persamaan(4)dan(5)log(2y)=logx×log(2z)log2log(xz)=logx×log(2z)log2log2(logxlogz)=logx×log(2z)log2×logxlog2×logz=logx×(log2+logz)log2×logxlog2×logz=logx×log2+logx×logzlog2×logz=logx×logzlog21=logx12=x.....(6)

.persamaan(2)Menentukan nilaizlog2y×log(4x)=logx×log(8yz)log2y×log(4(2yz))=logx×log(8yz)log2y×log(8yz)=logx×log(8yz)log(2y)=logx2y=xy=12x=12×12=14.....(7)x=2yz12=2(14)z1=z

.maka nilai untukxy5zadalahxy5z=(12).(14)5.1=12×45=12×(22)5=121+10=1211=12111=12pq{p=11q=1dan jelas bahwapdanqsaling primaJadi,p+q=11+1=12


DAFTAR PUSTAKA

  1. Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
  2. Sembiring, S. 2002. Olimpiade Matematika untuk SMU. Bandung: YRAMA WIDYA.