# Log. functions - comparing solutions, which is best?

1. Jul 3, 2013

### Tyrannosaurus_

1. The problem statement, all variables and given/known data
I'm not sure what my error is. Both solutions cannot be true.

2. Relevant equations
evaluating logarithms,
power of logarithms?

3. The attempt at a solution

SOLUTION ONE

=logxxn
=n(logxx)
=n(1)
=n

SOLUTION TWO

=logxxn
=(logxx)n
=(1)n
=1

Last edited: Jul 3, 2013
2. Jul 3, 2013

### Staff: Mentor

the logx x^n means logx(x^n) not logx(x)^n so your solution two is wrong.

the logx(x^n) is the same as n * logx(x)

3. Jul 3, 2013

### CAF123

Let $y = \log_x x^n$. Then $x^y = x^n \Rightarrow y = n$, so your solution one is correct.
$x^n$ is in the argument of the log, the log is not to the nth power.

4. Jul 3, 2013

### chinye11

The other 2 posts are correct: