A problem involving logarithms

[sahilmm15]

Homework Statement

$$\{ (log_2 9)^2 \}^{\frac {1} {log_2(log_2 9)} } \cdot (\sqrt 7)^\frac {1} {log_4 7}$$

Relevant Equations

$$ x^{log_a x}=a $$

This is my work. I don't know whether this is correct or not. What do you think of the problem.
Thanks!

 

[berkeman]

It works much better if you type your work into the forum using LaTeX. It is much easier for us to read and to quote any parts that we want to clarify/correct.

Please look at the LaTeX Guide in the lower left of the Edit window. Thank you.

 

[sahilmm15]

It works much better if you type your work into the forum using LaTeX. It is much easier for us to read and to quote any parts that we want to clarify/correct.

Please look at the LaTeX Guide in the lower left of the Edit window. Thank you.

Ya that's what I am finding difficult. I would edit it and post in latex form soon.

 

[sahilmm15]

After full ##1## hour of grind I finally did it. It was ridiculously tough for me. I think I am now some 3 percent familiar with it , lol.

 

[SammyS]

It works much better if you type your work into the forum using LaTeX. It is much easier for us to read and to quote any parts that we want to clarify/correct.

Please look at the LaTeX Guide in the lower left of the Edit window. Thank you.

After full ##1## hour of grind I finally did it. It was ridiculously tough for me. I think I am now some ##3## percent## familiar with it , lol.

A snip of your posted image gives the expression you are trying to simplify as:

By The Way: The ##{LaTeX}## code for ##\log_2 9## is ##\log_2 9## .

From your posted image, it appears that entire expression, ##\dfrac {1} {\log_2(\log_2 9)} \cdot
(\sqrt 7)^\left ( \dfrac {1} {\log_4 7} \right)## is to be used as the exponent for ##(\log_2 9)^2 ## .

You did correctly find that ## (\sqrt 7)^\left ( \dfrac {1} {\log_4 7} \right) = 2## .

But there are mistakes elsewhere.

 

[abcd112358]

I see two mistakes right now.

1. You left out the factor of 2 you derived in your step number 5

2. You forgot to set one bracket:
Your step from $$x^{2^{log_x(2)}}$$ to $$x\cdot x^{log_x(2)}$$ is wrong. There need to be brackets around both x: $$(x\cdot x)^{log_x(2)}$$ Therefore the exponent needs to be applied to both x $$x^{log_x(2)} \cdot x^{log_x(2)}$$

I think with these informations you should be able to find the correct answere. But feel free to ask additional questions.

 

[vela]

Relevant Equations:: $$ x^{log_a x}=a $$

That's not correct. For instance, if ##x=16## and ##a=2##, you get
$$16^{\log_2 16} = 16^4 = 65536 \ne 2.$$

 

[SammyS]

You did correctly find that ## (\sqrt 7)^\left ( \dfrac {1} {\log_4 7} \right) ## ##= 2## .

I don't think I've even seen ##\large{\log_{A^F} C^F= \log_{A} C }## stated as a rule for logarithms, but it does work.

(In reference to you changing ##\large{\log_{(\sqrt{7})^2} 2^2}## to ## \log_{(\sqrt{7})} 2 ## .)

 

[Mark44]

Relevant Equations:: $$ x^{log_a x}=a $$

That's not correct. For instance, if ##x=16## and ##a=2##, you get
$$16^{\log_2 16} = 16^4 = 65536 \ne 2.$$

But ##a^{\log_a x} = x## is the corrected version.

 

[aheight]

This is my work. I don't know whether this is correct or not. What do you think of the problem.

I think it would be helpful if you divide and conquer, do each part separately, and reduce everything to logarithms base "e" using the formula ##a^b=e^{b\ln a}##. Take for example the expression ##\log_2(9)##: Let ##u=\log_2(9)##. That means ##2^u=9## or ##e^{u\ln(2)}=9##. Thus ##\log_2(9)=\frac{\ln(9)}{\ln(2)}##. Do every piece separately this way and then put all the pieces together.