Question

Evaluate the logorithm
log(6)1/36

Answer 1

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:
\[\frac{ 1 }{ a^m }=a ^{-m}\]
\[\log_{a} b^n \iff (n)\log_{a} b\]

Answer 2

So 6 is a, m is 2, b is 1/36? And then I just plug them in? @Owlcoffee

Answer 3

remember I transformed the 36 into 6^2.

Answer 4

ok so it would be 1/6^2 = 6^-2? @Owlcoffee

Answer 5

That is correct.

Answer 6

How do I solve that or is that the answer? @Owlcoffee

Answer 7

So far you have
\[\log_{6} 6^{-2}\]
You can use the other property of logarithms I told you:
\[\log_{a} b^n \iff (n)\log_{a} b\]

Answer 8

Which number is n? @Owlcoffee

Answer 9

Observe the structure
"n" is the exponent above "6"

Answer 10

ok so -2. So it would be log(6)36^-2 <----->(-2)log(6) 36 @Owlcoffee

Answer 11

That is not correct, don't forget that it has to be strictly the 6^-2.
6... not 36.

Answer 12

oh jeez right. Ok log(6)6^-2 <---->(-2)log(6)6 @Owlcoffee

Answer 13

That is correct.
Now, what is \(\log_{6} 6\) equal to?

Answer 14

-2? @Owlcoffee

Answer 15

That's incorrect.
To what exponent do you have to raise "6" in order to obtain as a result "6"?

Answer 16

1? I'm sorry i'm just really bad at math. it might take me a while....

Answer 17

That is correct \[\log_{6} 6=1\]

Answer 18

Yes! THANK YOU!

Answer 19

So, what would be the result of this?
\[(-2)\log_{6} 6\]

Answer 20

-2 right?

Answer 21

Correct.

Answer 22

Yay! I really appreciate your help amd patience. Like this helped sooo much.

Answer 23

No problem, that's why we're helpers here.