3 szkoły ponadpodstawowej

III liceum

III technikum

MATeMAtyka 3. Zakres podstawowy

a) Z tablic logarytmów odczytujemy, że: 0,7168≈log5,21   Zatem: 100,7168≈10log5,21=5,21   <hr> b) Z tablic logarytmów odczytujemy, że: 0,7348≈log5,43   Zatem: 100,7348≈10log5,43=5,43   <hr> c) Z tablic logarytmów odczytujemy, że: 0,1106≈log1,29   Zatem: 100,1106≈10log1,29=1,29   <hr> d) Z tablic logarytmów odczytujemy, że: 0,9777≈log9,5   Zatem: 100,9777≈10log9,5=9,5   <hr> e) Z tablic logarytmów odczytujemy, że: 0,8451≈log7   Zatem: 100,8451≈10log7=7   <hr> f) Z tablic logarytmów odczytujemy, że: 0,9956≈log9,9   Zatem: 100,9956≈10log9,9=9,9

Rozwiązanie 1

a) 2,43⋅3,74=10log2,43⋅10log3,74≈100,3856⋅100,5729=100,3856+0,5729=100,9585≈10log9,09=9,09

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M834 80h400000v40h-400000z'/></svg>​log105​=    =log(1031​105​)−log1031​5​=   =log105−31​−31​5​=

<table style="height:112px;width:1109px;border-collapse:collapse;background-color:#dcdcdc;" border="0"> <tbody> <tr style="height:104px;"> <td style="width:1104.38px;height:104px;"> Niech a&gt;0, b&gt;0 oraz a≠1. Logarytmem o podstawie a z liczby b nazywamy wykładnik potęgi, do której należy podnieść podstawę a, by otrzymać liczbę logarytmowaną b. loga​b=c, gdy ac=b &nbsp; &nbsp; Z powyższej definicji wynika następująca własność, z której będziemy korzystać: aloga​b=b &nbsp; </td> </tr> </tbody> </table> <hr> a) 10log0,01=0,01 &nbsp;

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