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Mathematics 17 Online
OpenStudy (anonymous):

log 0.06

OpenStudy (zehanz):

If you write it as \(\log\dfrac{6}{100} \), you can use the rule: \(\log\dfrac{a}{b}=\log a - \log b\).

OpenStudy (zehanz):

But if you only need to know an approximation, you can type it into a calculator, of course...

OpenStudy (anonymous):

no i have to do it without a calculator and have no idea how!

OpenStudy (anonymous):

so can i just put log3-log50?

OpenStudy (zehanz):

You could use the method I described above. One of the two logs can be calculated, the other one stays as it is...

OpenStudy (anonymous):

log3-log50 which one stays as it is though?

OpenStudy (zehanz):

No, although 6/100=3/50, that is not convenient here. I would write it as: \(\log6 - \log 100\), then I can see what log 100 is...can you?

OpenStudy (anonymous):

what happened here? http://openstudy.com/users/hope_nicole#/updates/51475e90e4b04cdfc5831947

OpenStudy (zehanz):

Mine seems like a "second opinion" :) It is not different than @dpaInc's...

OpenStudy (anonymous):

second time around, it prolly be correct... :)

OpenStudy (anonymous):

@dpaInc , i submitted that and she said it was not right.

OpenStudy (zehanz):

Well, is there an explanation given then? Are there rules you have to apply?

OpenStudy (zehanz):

You could write log 6 as log 2 + log 3, but it doesn't get simpler if you do.

OpenStudy (anonymous):

you can simplify (or use the log rules again) log(6) = log(3*2) = ???

OpenStudy (anonymous):

i dont think so, i will go back and check though to make sure! isnt log 100 =2?

OpenStudy (zehanz):

It is!

OpenStudy (anonymous):

okay, this is exactly what i sent her(i ad to show work) log(0.06)=log(6/100)=log(6)−log(100)=log(6)−2

OpenStudy (zehanz):

That is what we came up with. The only thing you could do further, is: log(0.06)=log(6/100)=log(6)−log(100)=log(2*3)−2=log(2)+log(3)-2

OpenStudy (anonymous):

thanks guys, i will try that!

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