Thread: Solve for X... I Think I Did...?

The question Logx(X^5) = 5

The first x is sub-script.

I have:

1) Logx(X^5) = 5
2) X^5 = X^5
3) x = x

I imagine I'm wrong, as it seemed to easy. Where, if I did, go wrong? Thanks.

Originally Posted by iwanttogotouni

The question Logx(X^5) = 5

The first x is sub-script.

I have:

1) Logx(X^5) = 5
2) X^5 = X^5
3) x = x

I imagine I'm wrong, as it seemed to easy. Where, if I did, go wrong? Thanks.

Does $\displaystyle x = X$?

If so then answer number one is correct and I'd not write it any more than that. The rules explaining this are below

$\displaystyle \log_a(a) = 1$

$\displaystyle \log_a(a^k) = k\,log_a(a) = k$

So:

x = 1?

And was "Does x = X," a rhetorical question, or are you asking about the difference of the two x's?

Edit: Is k in the second equation, or do they just look together because they jumbled up?

Originally Posted by iwanttogotouni

So:

x = 1?

And was "Does x = X," a rhetorical question, or are you asking about the difference of the two x's?

No, x is not 1. This is because of domain restrictions on the logarithm's base

$\displaystyle log_a(b) = c$ is equivalent to $\displaystyle b = a^c$

$\displaystyle a$ cannot be one because 1 raised to any power is always 1


There are infinitely many values of x.

$\displaystyle x \in \mathbb{R} \, \, , \: x > 0 \, \, , \, \, x \neq 1$

No doubt someone else knows better notation than me for that - what it says is that x can be any real number greater than 0 which is not 1.

$\displaystyle x = \sqrt{e \pi}$ is a valid solution for example


edit: about the X and x I was asking because they are generally taken as different variables although if they were different there'd be no way of solving the equation.

edit 2: the k is just being used as an example to demonstrate the log power law, it is unrelated to the question you ask. k is often used to denote a constant

two methods I know:
Method 1
log(x^5) = 5
x^(log(x^5) = x^5
x^5 = x^5 which is true for all x

Method 2
log(x^5) = 5
5log(x) = 5
5=5 which also is true (seems like an ugly "proof"? what can I do in order to make it look more sweet?)

Oh, crap, I read it wrong. You mean step one is right? When you said answer, I kind of differentiated.

Yeah, step 1 is fine. It's a common result so would require no further working in a question

Ok, thank-you very much. Sorry, I can be a literal person at times; I often over-analyze things.