Logarithm

[Merida]

Determine the sum of all real value of x such that Log x ^ log x = ( log ₓ 10 ) ³
I understood, we must use change base equation , but I don’t know how to proceed further

 

[Dr.Peterson]

First, we need to be sure of the problem. Is it this?

[MATH](\log(x))^{\log(x)}=(\log_x(10))^3[/MATH]​

And is the log on the left-hand side the common (base 10) log?

If so, apply the change of base formula to [MATH]\log_x(10)[/MATH] and rearrange the RHS to look like the LHS.

 

[Merida]

Yes that’s the question

 

[Dr.Peterson]

Yes that’s the question

Now do what I suggested, and show what you get.

 

[Merida]

This is my solution and I got just one value for x as 1/1000 , is that right ?

 

[pka]

Yes that’s the question

Are you sure that it is not \(\large\log\left(x^{log(x)}\right)=\left(\log_x(10)\right)^3\)

 

[Merida]

Are you sure that it is not \(\large\log\left(x^{log(x)}\right)=\left(\log_x(10)\right)^3\)

Yup sure

 

[Dr.Peterson]

This is my solution and I got just one value for x as 1/1000 , is that right ?

Yes, that's right.

Here's how I did it, similar to yours, which was why I was confident of my interpretation:

[MATH](\log(x))^{\log(x)}=(\log_x(10))^3[/MATH] becomes [MATH](\log(x))^{\log(x)}=\left(\frac{1}{\log(x)}\right)^3 = \left(\log(x)\right)^{-3}[/MATH]
and setting the exponents equal, [MATH]\log(x) = -3[/MATH], so that [MATH]x = 10^{-3}[/MATH].