A, let's knock this out.
Ili kutatua swali hili, tutatumia kanuni ya kubadilisha 'base' ya logariti, ambayo inasema logba=logbloga, ambapo log bila 'base' maalum inamaanisha 'base' 10. Pia tutatumia kanuni za logariti kama vile log(xy)=logx+logy na log(xn)=nlogx.
Tunapewa:
log2=0.36103
log3=0.47712
log5=0.69897
log7=0.8451
a) log2(18)
Step 1: Badilisha 'base' ya logariti.
log2(18)=log2log18
Step 2: Rahisisha log18 kwa kutumia kanuni za logariti.
log18=log(2×32)=log2+log(32)=log2+2log3
Step 3: Badilisha thamani ulizopewa na ukokotoe.
log18=0.36103+2(0.47712)=0.36103+0.95424=1.31527
log2(18)=0.361031.31527≈3.6431
Step 4: Jibu la mwisho.
3.6431
b) log15(45)
Step 1: Badilisha 'base' ya logariti.
log15(45)=log15log45
Step 2: Rahisisha log45 na log15 kwa kutumia kanuni za logariti.
log45=log(32×5)=2log3+log5
log15=log(3×5)=log3+log5
Step 3: Badilisha thamani ulizopewa na ukokotoe.
log45=2(0.47712)+0.69897=0.95424+0.69897=1.65321
log15=0.47712+0.69897=1.17609
log15(45)=1.176091.65321≈1.4057
Step 4: Jibu la mwisho.
1.4057
c) log7(200)
Step 1: Badilisha 'base' ya logariti.
log7(200)=log7log200
Step 2: Rahisisha log200 kwa kutumia kanuni za logariti.
log200=log(23×52)=3log2+2log5
Step 3: Badilisha thamani ulizopewa na ukokotoe.
log200=3(0.36103)+2(0.69897)=1.08309+1.39794=2.48103
log7(200)=0.84512.48103≈2.9358
Step 4: Jibu la mwisho.
2.9358
d) log3(27)
Step 1: Andika 27 kama 'power' ya 3.
27=33
Step 2: Tumia kanuni ya logariti logb(bn)=n.
log3(27)=log3(33)=3
Step 3: Jibu la mwisho.
3
e) log4(20)
Step 1: Badilisha 'base' ya logariti.
log4(20)=log4log20
Step 2: Rahisisha log20 na log4 kwa kutumia kanuni za logariti.
log20=log(22×5)=2log2+log5
log4=log(22)=2log2
Step 3: Badilisha thamani ulizopewa na ukokotoe.
log20=2(0.36103)+0.69897=0.72206+0.69897=1.42103
log4=2(0.36103)=0.72206
log4(20)=0.722061.42103≈1.9679
Step 4: Jibu la mwisho.
1.9679
f) log4(64)
Step 1: Andika 64 kama 'power' ya 4.
64=43
Step 2: Tumia kanuni ya logariti logb(bn)=n.
log4(64)=log4(43)=3
Step 3: Jibu la mwisho.
3
Tuma swali linalofuata 📸