Logarithm confused

[Loki123]

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?

 

[BigBeachBanana]

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

$$3^{\log_{3}x}=x$$

 

[Loki123]

$$3^{\log_{3}x}=x$$

what about that?

 

[BigBeachBanana]

what about that?

$$3^{\log_{3}(\log_{2}x-9)}=\log_{2}x-9$$Do the same on the RHS.

 

[JeffM]

Terrible notation.

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

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

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

$$log_2(x) - 9 = 9 * \left (1 - \dfrac{8}{log_2(x)} \right ) = 9 - \dfrac{72}{log_2(x)}.$$
Now what?

 

[Loki123]

Terrible notation.

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

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

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

$$log_2(x) - 9 = 9 * \left (1 - \dfrac{8}{log_2(x)} \right ) = 9 - \dfrac{72}{log_2(x)}.$$
Now what?

This:

Thank you for such detailed explanation. It really helped.

 

[JeffM]

This:
View attachment 31584
Thank you for such detailed explanation. It really helped.

I am glad the detail helped. Good work on recognizing the quadratic.

But you overlooked something.

$$log_3(log_2(2^6) - 9) = log_3(6 - 9) \text { TILT}.$$
You need to test your answers against the ORIGINAL equations.

 

[BigBeachBanana]

@Loki123,
Another way you can see how they got from line 2 to line 3 is:
$$\log_{3}(a)=\log_{3}(b)\implies a=b$$

 

[Loki123]

Ye

I am glad the detail helped. Good work on recognizing the quadratic.

But you overlooked something.

$$log_3(log_2(2^6) - 9) = log_3(6 - 9) \text { TILT}.$$
You need to test your answers against the ORIGINAL equations.

Yes I wrote that on the side. The condition is x>2^9 so the only correct answer is x=2^12

 

[Deleted member 4993]

Ye

Yes I wrote that on the side. The condition is x>2^9 so the only correct answer is x=2^12

I did not see that in your attachment.

 

[Loki123]

I did not see that in your attachment.

I didn't mean on the side in the picture I sent. Just on the side in general.

 

[Steven G]

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 $\displaystyle 2 + \log_3(1-4\log_x4) = \log_39+ \log_3(1-4\log_x4)=\log_3[9(1-4\log_x4)] =\log_3[9(1-\dfrac{8}{\log_2x})]$

 

[Steven G]

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 $\displaystyle 2 + \log_3(1-4\log_x4) = \log_39+ \log_3(1-4\log_x4)=\log_3[9(1-4\log_x4)] =\log_3[9(1-\dfrac{8}{\log_2x})]$

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

 

[Deleted member 4993]

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

No artificial light - it is just my "blinding brilliance".

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

 

[Steven G]

No artificial light - it is just my "blinding brilliance".

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

If you had blinding brilliance you wouldn't be in the corner!

 

[JeffM]

If you had blinding brilliance you wouldn't be in the corner!

Ahh no, Steven a person with corruscating brilliance may blind himself with his own dazzle. It happens to me all the time.