anonymous
  • anonymous
3. Evaluate: Log4 1/64
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Does anyone know how to do this?
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle \log_{a}(a)=1 }\) (Rule, for a>0)
anonymous
  • anonymous
huh?

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SolomonZelman
  • SolomonZelman
That was a rule that we'll apply..
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle \frac{1}{64}=\frac{1}{4^4}=4^{-4} }\)
anonymous
  • anonymous
i dont get how to read that?
SolomonZelman
  • SolomonZelman
That rule said, "log with base "a", of "a" is equivalent to 1" (this is true for every value of a that is bigger than 0) For example \(\large\color{#000000 }{ \displaystyle \log_2(2)=1 }\)
SolomonZelman
  • SolomonZelman
that would read as, "log base 2, of 2 is equivalent to 1"
anonymous
  • anonymous
log_4(1)=64?
SolomonZelman
  • SolomonZelman
oh (1) in that rule, a\(\ne\)1 (2) another rule \(\log_b(1)=0\) regardless of the base (b)
anonymous
  • anonymous
im lost how would it be then?
SolomonZelman
  • SolomonZelman
■ Rule 1 \(\large\color{#000000 }{ \displaystyle \log_a(b^c)=c\log_a(b) }\) (exponent inside the logarithm would go outside the logarithm) ■ Rule 2 \(\large\color{#000000 }{ \displaystyle \log_a(a)=1}\) (for all a, except a=1) ■ Rule 3 \(\large\color{#000000 }{ \displaystyle \log_a(1)=0 }\) (for positive bases (except a=1), and YOU DON'T NEED THAT NOW )
SolomonZelman
  • SolomonZelman
I have written this previously, \(\large\color{#000000 }{ \displaystyle (1/64)=4^{-4} }\) Do you agree with that statement?
SolomonZelman
  • SolomonZelman
I will put up an example of a similar problem while you are going over what I said...
anonymous
  • anonymous
\[\frac{1}{64}=\frac{1}{4^3} \]
SolomonZelman
  • SolomonZelman
yes, thanks, I made a mistake ...
SolomonZelman
  • SolomonZelman
\(\color{#000000 }{ \displaystyle \bf Example~~problem: }\) Evaluate: \(\color{#000000 }{ \displaystyle \log_5\left(\frac{1}{25}\right) }\) \(\color{#000000 }{ \displaystyle \bf Example~~solution: }\) Applying the rules of exponents, I get: \(\tiny{\\[1.5em]}\) \(\color{#000000 }{ \displaystyle \frac{1}{25}=\frac{1}{5^2} }\) (because \(25=5^2\)) \(\color{#000000 }{ \displaystyle \frac{1}{5^2}=5^{-2} }\) So, we can re-write the logarithm as, \(\color{#000000 }{ \displaystyle \log_5\left(\frac{1}{25}\right)\Longrightarrow\log_5\left(5^{-2}\right) }\) And we can use the [1] \(\color{#000000 }{ \displaystyle \log_5\left(\frac{1}{25}\right)\Longrightarrow(-2)\log_5\left(5\right) }\) And we can use [3] \(\color{#000000 }{ \displaystyle (-2)\log_5\left(5\right) \Longrightarrow~~~=-2\cdot 1 =-2 }\) References, [1] \(\color{#000000 }{ \displaystyle \log_a(b^c)=c\log_a(b) }\) (exponent inside the logarithm would go outside the logarithm) [3] \(\color{#000000 }{ \displaystyle \log_a(a)=1}\) (for all a, except a=1)

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