[math]3^{\log_{3}x}=x[/math]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? View attachment 31583
what about that?[math]3^{\log_{3}x}=x[/math]
[math]3^{\log_{3}(\log_{2}x-9)}=\log_{2}x-9[/math]Do the same on the RHS.what about that?
This: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?
I am glad the detail helped. Good work on recognizing the quadratic.
Yes I wrote that on the side. The condition is x>2^9 so the only correct answer is x=2^12I 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.
I did not see that in your attachment.Ye
Yes I wrote that on the side. The condition is x>2^9 so the only correct answer is x=2^12
I didn't mean on the side in the picture I sent. Just on the side in general.I did not see that in your attachment.
Dr Subotosh Khan,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 \(\displaystyle 2 + \log_3(1-4\log_x4) = \log_39+ \log_3(1-4\log_x4)=\log_3[9(1-4\log_x4)] =\log_3[9(1-\dfrac{8}{\log_2x})] \)
No artificial light - it is just my "blinding brilliance".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
If you had blinding brilliance you wouldn't be in the corner!No artificial light - it is just my "blinding brilliance".
I don't need no stinking external light from outside ..............
Ahh no, Steven a person with corruscating brilliance may blind himself with his own dazzle. It happens to me all the time.If you had blinding brilliance you wouldn't be in the corner!