Question

3. Evaluate: Log4 1/64

Answer 1

Does anyone know how to do this?

Answer 2

\(\large\color{#000000 }{ \displaystyle \log_{a}(a)=1 }\) (Rule, for a>0)

Answer 3

huh?

Answer 4

That was a rule that we'll apply..

Answer 5

\(\large\color{#000000 }{ \displaystyle \frac{1}{64}=\frac{1}{4^4}=4^{-4} }\)

Answer 6

i dont get how to read that?

Answer 7

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
\(\large\color{#000000 }{ \displaystyle \log_2(2)=1 }\)

Answer 8

that would read as,
"log base 2, of 2 is equivalent to 1"

Answer 9

log_4(1)=64?

Answer 10

oh
(1) in that rule, a\(\ne\)1
(2) another rule \(\log_b(1)=0\) regardless of the base (b)

Answer 11

im lost how would it be then?

Answer 12

■ 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 )

Answer 13

I have written this previously,
\(\large\color{#000000 }{ \displaystyle (1/64)=4^{-4} }\)

Do you agree with that statement?

Answer 14

I will put up an example of a similar problem while you are going over what I said...

Answer 15

\[\frac{1}{64}=\frac{1}{4^3} \]

Answer 16

yes, thanks, I made a mistake ...

Answer 17

\(\color{#000000 }{ \displaystyle \bf Example~~problem: }\)
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)