logarithm help pls

[Shoppingal]

question asks solve for x

this is what I have done

logxX^5=5

x^5=x^5

thanks for any help

 

[mmm4444bot]

solve for x

this is what I have done

logxX^5=5

Is this what they gave you?

\(\displaystyle \log_x x^{5} = 5\)

 

[Shoppingal]

yes

 

[mmm4444bot]

You previously wrote that the equation is something that you did, so I was not sure.

Please tell us what you think the meaning is for the following generic symbol.

\(\displaystyle \log_b \; x\)

---

Your exercise requires no calculations. If you understand the meaning of the symbol above (i.e., if you understand the very definition of a logarithm), then you should be able to answer your exercise simply by looking at it and thinking about the meaning.

Oh, you also need to know about the domain.

 

[HallsofIvy]

\(\displaystyle log_x x^5= 5\) or, more generally, \(\displaystyle log_x x^a= a\) for any a, is true for any positive x.

The logarithm, base a, \(\displaystyle log_a(x)\), can be defined as the inverse function to \(\displaystyle f(x)= a^x\). \(\displaystyle log_a(a^x)= f^{-1}(f(x))= x\) by definition of "inverse" function. In particular, replacing "x" in that last equation by 5 and replacing "a" by x, \(\displaystyle log_x(x^5)= 5\).

 

[mmm4444bot]

x^5=x^5

Perhaps, I do not understand your style of communication, Shoppingal.

When you wrote that you did this:

x^5 = x^5

was that your way of saying that you're thinking the given equation is true for any value of x?

If I were sitting next to you, I would often be asking you what you're thinking. It's hard to do that via bulletin boards.

 

[lookagain]

\(\displaystyle log_x x^5= 5\) or, more generally, \(\displaystyle log_x x^a= a\) for any a, is true for any positive x.

x cannot equal 1.


One is not a legitimate base for logarithms.


\(\displaystyle \log_1 (1^a) \ \ = \ \ a*\log_11 \ \ is \ \ indeterminate, \ \ \ as \ \ \log_11 \ \ is \ \ indeterminate.\)


1 raised to any real number equals 1.

 

[mmm4444bot]

Good catch, lookagain. I probably would not have received full credit.

I initially felt that this exercise was little more than a trivial tester of basic-logarithmic definition (i.e., logarithms are exponents), but now I see that my former perception (of triviality) is flawed.