Log Problem: deriv of f(x) = log_2(log_3(log_4(x)))

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mikexz

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I am trying to do this question, find the derivative

f(x) = log[sub:17ully7v]2[/sub:17ully7v] (log[sub:17ully7v]3[/sub:17ully7v]( log[sub:17ully7v]4[/sub:17ully7v] x)

My attempt,
I reversed it to

4^3^2^y

but I am not sure whether I can use this property (x^a)^b = a^ab because if I were to do this
(2^3)^4 = x 2^(3^4) = c where c does not equal x.
 
Re: Log Problem

What are you trying to DO with it? A problem statement might help.

If \(\displaystyle y = log_{2}log_{3}log_{4}(x)\), then

\(\displaystyle 2^{y} = log_{3}log_{4}(x)\), then

\(\displaystyle 3^{2^{y}} = log_{4}(x)\), then

\(\displaystyle 4^{3^{2^{y}}} = x\)

Is that what is wanted? It seems pretty straight-forward.
 
Re: Log Problem

thanks lol, I forgot, I was trying to get the derivative
 
mikexz said:
I am trying to do this question, find the derivative

f(x) = log[sub:1jtwtfzj]2[/sub:1jtwtfzj] (log[sub:1jtwtfzj]3[/sub:1jtwtfzj]( log[sub:1jtwtfzj]4[/sub:1jtwtfzj] x)

My attempt,
I reversed it to

4^3^2^y

but I am not sure whether I can use this property (x^a)^b = a^ab because if I were to do this
(2^3)^4 = x 2^(3^4) = c where c does not equal x.
Click to expand...

\(\displaystyle f(x) \, =\, \frac{1}{ln(2)}\cdot ln[\frac{1}{ln(3)}\cdot ln(\frac{ln(x)}{ln(4)})]\)

Now use chain rule - very carefully.....
 
That was what I was originally doing but just asked my teacher (via email) said that this way is better (I don't think it is valid though, is there a proof I can do to prove it wrong? or is it correct?)

4^3^2^y = x using (x^a)^b = x^(ab)
it becomes
x= 4^(6y)
then derive.

OR (this method from another student)

4^3^2^y = x
it becomes
x= (4^3^2)^y
x = 4096^y
then derive.

Is either of these solutions correct?? If not, why?
I think atleast one of them HAS to be wrong.

thanks for your help
 
mikexz said:
x= 4^(6y) then derive.

OR (this method from another student)

x = 4096^y then derive.
Click to expand...
What is 4[sup:2xfp48yc]6[/sup:2xfp48yc]? :wink:

Eliz.
 
thanks but

2^(3^4) doesn't equal (2^3)^4 ??? Why is that?

Also with that property does it work with more than 2 exponents so ((x^a)^b)^c = x^(abc) ?

-I am really rusty with my exponents, expecially the properties that I don't use often, is there a website that has ALL the properties? and maybe some material that would be useful for solving thinking/contest type questions?

thanks again
 
\(\displaystyle 2^{3^4}\, = \, 2^{81}\)

it is not

\(\displaystyle (2^3)^4\, = \, 2^{12} \, or \, = 8^4\)
 
mikexz said:
That was what I was originally doing but just asked my teacher (via email) said that this way is better (I don't think it is valid though, is there a proof I can do to prove it wrong? or is it correct?)

4^3^2^y = x using (x^a)^b = x^(ab)
it becomes
x= 4^(6y) <<<< This is wrong - see my example above
then derive.

OR (this method from another student)

4^3^2^y = x
it becomes
x= (4^3^2)^y
x = 4096^y<<<< This is wrong too - see my example above
then derive.

Is either of these solutions correct?? If not, why?
I think atleast one of them HAS to be wrong.

thanks for your help
Click to expand...
 
Hello, Mike!

Find the derivative: . \(\displaystyle f(x) \:= \:\log_2[\log_3(\log_4 x)]\)
Click to expand...

I used the Base Change formula . . .

\(\displaystyle f(x) \;=\;\log_2\left[\log_3\left(\frac{\ln x}{\ln 4}\right)\right] \;=\;\log_2\bigg[\log_3(\ln x) - \log_3(\ln4)\bigg] \;= \;\log_2\left[\frac{\ln(\ln x)}{\ln3} - \frac{\ln(\ln 4)}{\ln 3}\right]\)

\(\displaystyle f(x) \;=\; \log_2\left[\frac{\ln(\ln x) - \ln(\ln4)}{\ln 3}\right] \;=\; \log_2\bigg[\ln(\ln x) - \ln(\ln4)\bigg] - \log_2(\ln 3)\)

\(\displaystyle f(x) \;=\; \frac{\ln\left[\ln(\ln x) - \ln(\ln4)\right]}{\ln 2} - \frac{\ln(\ln3)}{\ln 2} \;=\;\frac{1}{\ln2}\bigg\{\ln\bigg[\ln(\ln x) - \ln(\ln 4)\bigg] - \ln(\ln 3)\bigg\}\)


\(\displaystyle \text{Then: }\;f'(x) \;=\;\frac{1}{\ln 2}\cdot\frac{1}{\ln(\ln x)-\ln(\ln4)} \cdot\frac{1}{\ln x}\cdot\frac{1}{x}\)

\(\displaystyle \text{Therefore: }\;f'(x)\;=\;\frac{1}{\ln2\cdot x\cdot \ln x\cdot\left[\ln(\ln x)-\ln(\ln4)\right]}\)


 
Thanks :D

why is 2^3^4 = 2^(3^4) ? Is there some explanation of this concept on some website (explanation of all possible scenarios with exponents)? because I really want to fully understand this before the math test in 2 days.

thank you so much
 
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