anonymous
  • anonymous
Evaluate the logorithm log(6)1/36
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Owlcoffee
  • Owlcoffee
I would recommend effectuating the corresponding exponential transformations and trying to form a base which respects the one in the logarithm. Since the level is labeled "high" I suppose you have no problem with that, so, just stating a possible method to evaluate the logarithm without calculator: \[36=6^2\] \[\log_{6} \frac{ 1 }{ 36 } \iff \log_{6} \frac{ 1 }{ 6^2 }\] With that transformed, I suggest you use the exponential property of negative exponents and bases on denominator, to posterior use the logarithmic property of exponential logarithmand. Here they are: \[\frac{ 1 }{ a^m }=a ^{-m}\] \[\log_{a} b^n \iff (n)\log_{a} b\]
anonymous
  • anonymous
So 6 is a, m is 2, b is 1/36? And then I just plug them in? @Owlcoffee
Owlcoffee
  • Owlcoffee
remember I transformed the 36 into 6^2.

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anonymous
  • anonymous
ok so it would be 1/6^2 = 6^-2? @Owlcoffee
Owlcoffee
  • Owlcoffee
That is correct.
anonymous
  • anonymous
How do I solve that or is that the answer? @Owlcoffee
Owlcoffee
  • Owlcoffee
So far you have \[\log_{6} 6^{-2}\] You can use the other property of logarithms I told you: \[\log_{a} b^n \iff (n)\log_{a} b\]
anonymous
  • anonymous
Which number is n? @Owlcoffee
Owlcoffee
  • Owlcoffee
Observe the structure "n" is the exponent above "6"
anonymous
  • anonymous
ok so -2. So it would be log(6)36^-2 <----->(-2)log(6) 36 @Owlcoffee
Owlcoffee
  • Owlcoffee
That is not correct, don't forget that it has to be strictly the 6^-2. 6... not 36.
anonymous
  • anonymous
oh jeez right. Ok log(6)6^-2 <---->(-2)log(6)6 @Owlcoffee
Owlcoffee
  • Owlcoffee
That is correct. Now, what is \(\log_{6} 6\) equal to?
anonymous
  • anonymous
-2? @Owlcoffee
Owlcoffee
  • Owlcoffee
That's incorrect. To what exponent do you have to raise "6" in order to obtain as a result "6"?
anonymous
  • anonymous
1? I'm sorry i'm just really bad at math. it might take me a while....
Owlcoffee
  • Owlcoffee
That is correct \[\log_{6} 6=1\]
anonymous
  • anonymous
Yes! THANK YOU!
Owlcoffee
  • Owlcoffee
So, what would be the result of this? \[(-2)\log_{6} 6\]
anonymous
  • anonymous
-2 right?
Owlcoffee
  • Owlcoffee
Correct.
anonymous
  • anonymous
Yay! I really appreciate your help amd patience. Like this helped sooo much.
Owlcoffee
  • Owlcoffee
No problem, that's why we're helpers here.

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