Logarithm confused

L

Loki123

Full Member
Joined
Sep 22, 2021
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790
I don't have the original problem.
The first line is the problem. I don't understand how they got second and third line. How do I get this? IMG_20220308_203751.jpg
 
Terrible notation.

[math]log_3\{log_2(x) - 9\} = 2 + log_3\{1 - 4log_x(4)\} = 2log_3(3) +log_3\{1 - 4log_x(4)\}.[/math]
Follow that? Now exponents.

[math]log_3\{log_2(x) - 9\} = 2log_3(3) +log_3\{1 - 4log_x(4)\} =\\ log_3(3^2) + log_3\{1 - log_x(4^4)\} = log_3(9) + log_3\{1 - log_x(2^8)\}.[/math]
Now change of base formula

[math]log_3\{log_2(x) - 9\} = log_3(9) + log_3\{1 - log_x(2^8)\} =\\ log_3(9) + log_3 \left \{ 1 - \dfrac{log_2(2^8)}{log_2(x)} \right \} = log_3(9) + log_3 \left \{ 1 - \dfrac{8}{log_2(x)} \right \} =\\ log_3 \left \{ 9 * \left ( 1 - \dfrac{8}{log_2(x)} \right ) \right \} .[/math]
Now we can drop the logs to the base 3.

[math]log_2(x) - 9 = 9 * \left (1 - \dfrac{8}{log_2(x)} \right ) = 9 - \dfrac{72}{log_2(x)}.[/math]
Now what?
 
JeffM said:
Terrible notation.

[math]log_3\{log_2(x) - 9\} = 2 + log_3\{1 - 4log_x(4)\} = 2log_3(3) +log_3\{1 - 4log_x(4)\}.[/math]
Follow that? Now exponents.

[math]log_3\{log_2(x) - 9\} = 2log_3(3) +log_3\{1 - 4log_x(4)\} =\\ log_3(3^2) + log_3\{1 - log_x(4^4)\} = log_3(9) + log_3\{1 - log_x(2^8)\}.[/math]
Now change of base formula

[math]log_3\{log_2(x) - 9\} = log_3(9) + log_3\{1 - log_x(2^8)\} =\\ log_3(9) + log_3 \left \{ 1 - \dfrac{log_2(2^8)}{log_2(x)} \right \} = log_3(9) + log_3 \left \{ 1 - \dfrac{8}{log_2(x)} \right \} =\\ log_3 \left \{ 9 * \left ( 1 - \dfrac{8}{log_2(x)} \right ) \right \} .[/math]
Now we can drop the logs to the base 3.

[math]log_2(x) - 9 = 9 * \left (1 - \dfrac{8}{log_2(x)} \right ) = 9 - \dfrac{72}{log_2(x)}.[/math]
Now what?
Click to expand...
This:
IMG_20220308_230222.jpg
Thank you for such detailed explanation. It really helped.
 
Ye
JeffM said:
I am glad the detail helped. Good work on recognizing the quadratic.

But you overlooked something.

[math]log_3(log_2(2^6) - 9) = log_3(6 - 9) \text { TILT}.[/math]
You need to test your answers against the ORIGINAL equations.
Click to expand...
Yes I wrote that on the side. The condition is x>2^9 so the only correct answer is x=2^12
 
Here is how I go from line 2 to line 3.
\(\displaystyle 2 + \log_3(1-4\log_x4) = \log_39+ \log_3(1-4\log_x4)=\log_3[9(1-4\log_x4)]\)

Remains to show that \(\displaystyle 4\log_x4=\dfrac{8}{\log_2x}\)
\(\displaystyle 4\log_x4= 4(\dfrac{\log_24}{log_2x}) = 4(\dfrac{2}{log_2x}) = \dfrac{8}{log_2x}\)

So \(\displaystyle 2 + \log_3(1-4\log_x4) = \log_39+ \log_3(1-4\log_x4)=\log_3[9(1-4\log_x4)] =\log_3[9(1-\dfrac{8}{\log_2x})] \)
 
Last edited:
Steven G said:
Here is how I go from line 2 to line 3.
\(\displaystyle 2 + \log_3(1-4\log_x4) = \log_39+ \log_3(1-4\log_x4)=\log_3[9(1-4\log_x4)]\)

Remains to show that \(\displaystyle 4\log_x4=\dfrac{8}{\log_2x}\)
\(\displaystyle 4\log_x4= 4(\dfrac{\log_24}{log_2x}) = 4(\dfrac{2}{log_2x}) = \dfrac{8}{log_2x}\)

So \(\displaystyle 2 + \log_3(1-4\log_x4) = \log_39+ \log_3(1-4\log_x4)=\log_3[9(1-4\log_x4)] =\log_3[9(1-\dfrac{8}{\log_2x})] \)
Click to expand...
Dr Subotosh Khan,
Do you have light in the corner? If so, you must extinguish it. You are to sit in a dark corner facing the wall after your last mistake.
Elite Member Steven
 
Steven G said:
Dr Subotosh Khan,
Do you have light in the corner? If so, you must extinguish it. You are to sit in a dark corner facing the wall after your last mistake.
Elite Member Steven
Click to expand...
No artificial light - it is just my "blinding brilliance".

I don't need no stinking external light from outside ..............
 
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