Logarithms, change of base: log4^2x^2-7log1/4|x|!=-2=0

[spacelights]

okay, here is the situation. I used to have a good mathematical brain but haven't turned it on for specific problematic functions in years. Recently, I was tested by a younger student to complete the following problem and I am having the darndest time finding the right criteria to give me a foundation of knowledge so I can work out the problem on my own. Below is the problem (at least from how I read it?):

log4² x² - 7 log¼ |x| ? -2 = 0

Can you please lead me into the right direction about how to solve???
Your help is most appreciated!!
Many thanks...R

 

[wjm11]

Re: Logarithms, change of base problem

log4² x² - 7 log¼ |x| ? -2 = 0

Hello, R,

I'm having trouble interpreting your problem statement. Are these logs supposed to be base 10 or base 4 or 1/4 or...? You indicated a change of base problem in your subject line; however, if the logs are the same base, and it's base 10, there is no need to change the base. If they are different bases, make them the same base (preferably 10 or e, so you can use your calculator). I assume you are familiar with the "change of base" formula for logarithms(?). There are many sites that list properties of logs. Here is one:

http://en.wikipedia.org/wiki/Logarithm

Also, is that a "is not equal to" sign to the left of -2? Can you clarify these points please?

 

[soroban]

Re: Logarithms, change of base problem

Hello, spacelights!

Your "equation" makes no sense . . .

\(\displaystyle \text{I }think\text{ I have the left side: }\:\log_{[4^2]}(x^2) - 7\log_{[\frac{1}{4}]}(x)\)

. . which is "not equal to two equals zero" . . . what?


\(\displaystyle \text{I would use this identity: }\:\log_a(b) \:=\:\frac{1}{\log_b(a)}\)


\(\displaystyle \text{The first term is: }\:\log_{[4^2]}(x^2) \;=\;2\log_{[4^2]}(x) \;=\;\frac{2}{\log_x(4^2)} \;=\;\frac{2}{2\log_x(4)} \;=\;\frac{1}{\log_x(4)}\)

\(\displaystyle \text{The second term is: }\:-7\log_{[\frac{1}{4}]}(x) \;=\;\frac{-7}{\log_x(\frac{1}{4})} \;=\;\frac{-7}{\log_x(4^{\text{-}1})} \;=\;\frac{-7}{-\log_x(4)} \;=\;\frac{7}{\log_x(4)}\)


\(\displaystyle \text{So the left side becomes: }\;\frac{1}{\log_x(4)} + \frac{7}{\log_x(4)} \;=\;\frac{8}{\log_x(4)}\)

. . Now what is the rest of the equation?

 

[spacelights]

To wjm11 and Soroban:

Both of your efforts are helpful. thank you. (I would have replied a lot sooner, thinking I would receive email notification if someone responded to my question.) The logarithm equation mystifies me, as well, and I don't seem to have all the character tools on my computer to write out such an equation. After studying the scraggly penmanship on my paper and, ruling out that a 'not equal to' -2 = 0, on the right side of the equation, is plain wrong, I will stab at what I believe is the logarithm equation written correctly (please bear-in-mind I will write out the symbols which my lack of computer knowledge restricts me from symbolizing properly): log 2 (base 4) x (squared) - 7 log (base 1/4) |x| -2 = 0 ; find x

Now that I am committing myself to an equation, I will do my best to solve it...I will also continue to check back regularly for any responses.

Thank you in advance,
R

 

[soroban]

Re: Logarithms, change of base problem

Hello, spacelights!

Ah, I think I understand . . .

\(\displaystyle \log_{[4^2]}(x^2) - 7\log_{[\frac{1}{4}]}(x) - 2 \:=\:0\)

\(\displaystyle \text{Base-change Formula: }\:\log_b(a) \:=\:\frac{\log a}{\log b}\)


\(\displaystyle \text{The first term is: }\:\log_{[4^2]}(x^2) \;=\;\frac{\log(x^2)}{\log(4^2)} \;=\;\frac{2\log(x)}{2\log(4)} \;=\;\frac{\log(x)}{\log(4)}\)

\(\displaystyle \text{The second term is: }\:-7\log_{[\frac{1}{4}]}(x) \;=\;-7\,\frac{\log(x)}{\log(\frac{1}{4})} \;=\;-7\frac{\log(x)}{\log(4^{-1})} \;=\;-7\,\frac{\log(x)}{-\log(4)} \;=\;\frac{7\log(x)}{\log(4)}\)


\(\displaystyle \text{So the equation becomes: }\;\frac{\log(x)}{\log(4)} + \frac{7\log(x)}{\log(4)} -2\;=\;0 \quad\Rightarrow\quad \frac{8\log(x)}{\log(4)} -2 \;=\;0\)

\(\displaystyle \text{We have: }\:\frac{8\log(x)}{\log(4)} \:=\:2 \quad\Rightarrow\quad 8\log(x) \:=\:2\log(4) \quad\Rightarrow\quad \log(x) \:=\:\tfrac{1}{4}\log(4)\)

. . \(\displaystyle \log(x) \:=\:\log\left(4^{\frac{1}{4}}\right) \quad\Rightarrow\quad x \:=\:4^{\frac{1}{4}} \:=\:\left(2^2\right)^{\frac{1}{4}} \;=\;2^{\frac{1}{2}}\)

\(\displaystyle \text{Therefore: }\;x \:=\:\sqrt{2}\)