Log question

[srayworth]

For x > 0 and x not equal to 1, log_16(x) =

A) 8 log_2(x)
B) 4 log_2(x)
C) 0.5 log_2(x)
D) 0.25 log_2(x)
E) 0.125 log_2(x)

The answer is D) however I’m not sure exactly why or how to get to that answer. I think I’m forgetting something simple about the basic rules of logs.
When I do this I change log_16(x) into 16^x
16^x=2^4x
2^4x = 4log_2(x)????
Thanks!

 

[tkhunny]

The first:

Go through the transformation: \(\displaystyle log_{a}(b)\;=\;\frac{log_{c}(a)}{log_{c}(b)}\) (for appropriate a, b, and c. I suggest using c = 2. Why would I suggest that?

 

[soroban]

Hello, srayworth!

Another approach . . .

\(\displaystyle \text{For }x > 0\text{ and }x \ne 1:\;\log_{16}(x)\: =\)

\(\displaystyle A)\;8\log_2(x) \quad B)\;4\log_2(x) \quad C)\;\tfrac{1}{2}\5\log_2(x) \quad D)\;\tfrac{1}{4}\log_2(x) \quad E)\;\tfrac{1}{8}\log_2(x)\)

\(\displaystyle \text{Let }\log_{16}(x) \:=\\)

\(\displaystyle \text{Then: }\:16^P \:=\:x \quad\Rightarrow\quad (2^4)^P \:=\:x \quad\Rightarrow\quad 2^{4P} \:=\:x\)

\(\displaystyle \text{Take logs, base 2: }\;\log_2(2^{4P}) \:=\:\log_2(x) \quad\Rightarrow\quad 4P\cdot\underbrace{\log_2(2)}_{\text{This is 1}} \:=\:\log_2(x)\)
. . . \(\displaystyle 4P \:=\:\log_2(x) \quad\Rightarrow\quad P \:=\:\tfrac{1}{4}\log_2(x)\)

\(\displaystyle \text{Therefore: }\:\log_{16}x \:=\:\tfrac{1}{4}\log_2(x) \;\;\hdots\;\text{ answer (D)}\)

 

[srayworth]

I see it now! Thanks very much!