• anonymous
plz prove the answer about laws of logarithms: 1.logb M=logb M+logb N 2.logb M/N=logb M-logb N 3.logb N exponent p=plogb N 4.logb exponent b=1 bN=N PLZ prove the answer give me the solution...thanks
  • Stacey Warren - Expert
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  • chestercat
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  • goformit100
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  • goformit100
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  • DebbieG
We aren't here to "prove the answer" and give you the solution :) but I'll help you get started on the first one. You're going to use the fact that \(\Large \log_{b}z=y \) means that \(\Large b^y=z\). It's important that you understand that relationship between log form and exponential form, so stop and stare at that for a bit if you need to! Now you want to show that \(\Large \log_{b}(MN)=log_{b}M+log_{b}N\). (That isn't quite what you have above, but that's what you meant, isn't it? :) So let's let \(\Large y=log_{b}M\) and \(\Large z=log_{b}N\) Then \(\Large b^y=M\) and \(\Large b^z=N\) (by changing from log to exponential form) This means that \(\Large MN= b^y\cdot b^z=b^{y+z}\) right? (by rules of exponents) So we have \(\Large MN= b^{y+z}\). Now convert that into log form and you have: \(\Large log_{b}(MN)=y+z\) But hey, look up there where we FIRST defined what y and z are...! Now sub those expressions for y and z back into that last equation above, and what do you get? That's the general idea... the others can be proven similarly. Just remember that a LOG is an EXPONENT, so the rules for working with exponents will apply. Once you switch form, it's usually easier to see that :)

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