Log. functions - comparing solutions, which is best?

[Tyrannosaurus_]

Homework Statement

I'm not sure what my error is. Both solutions cannot be true.

Homework Equations

evaluating logarithms,
power of logarithms?

The Attempt at a Solution

SOLUTION ONE

=log_xxⁿ
=n(log_xx)
=n(1)
=nSOLUTION TWO

=log_xxⁿ
=(log_xx)ⁿ
=(1)ⁿ
=1

Please help me understand what I've done wrong. Thanks!

 

[jedishrfu]

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)

 

[CAF123]

SOLUTION TWO

=log_xxⁿ
=(log_xx)ⁿ
=(1)ⁿ
=1

Please help me understand what I've done wrong. Thanks!

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.

 

[chinye11]

The other 2 posts are correct:

Your solution 1 uses log_x(xⁿ)
Your solution 2 uses (log_x(_x))ⁿ

An easy way to avoid this is to use brackets whenever applying a function: