1. ## properties of logarithms

I have the expression $4log_{3}\sqrt[2]{x}-4log_{3}x$. In simplifying the first term I need to get rid of the square root and change the coefficient of 4 to an exponent. So if I get rid of the square root first that would be $x^{1/2}$ so when I now change the coefficient of 4 to an exponent, shouldnt the term be $log_{3}(x^{1/2})^4$? It seems that the actual answer is $log_{3}x^{1/2(4)}$ but according to the properties of logarithms I dont understand why. The properties say that $\sqrt[x]{y}=y^{1/x}$ and $mlog n=logn^m$ so why is the simplification of the term $log_{3}x^{1/2(4)}$ and not $log_{3}(x^{1/2})^4$?

In other words if $mlog n=logn^m$ then why am I simply multiplying the exponent of 1/2 in $x^{1/2}$ by 4 instead of raising it to a power of 4?

I have the expression $4log_{3}\sqrt[2]{x}-4log_{3}x$. In simplifying the first term I need to get rid of the square root and change the coefficient of 4 to an exponent. So if I get rid of the square root first that would be $x^{1/2}$ so when I now change the coefficient of 4 to an exponent, shouldnt the term be $log_{3}(x^{1/2})^4$? It seems that the actual answer is $log_{3}x^{1/2(4)}$ but according to the properties of logarithms I dont understand why. The properties say that $\sqrt[x]{y}=y^{1/x}$ and $mlog n=logn^m$ so why is the simplification of the term $log_{3}x^{1/2(4)}$ and not $log_{3}(x^{1/2})^4$?

In other words if $mlog n=logn^m$ then why am I simply multiplying the exponent of 1/2 in $x^{1/2}$ by 4 instead of raising it to a power of 4?
Let $x>0$ and $a,b \in R$ , then :
$a log (x)^b = log \left( (x)^b \right)^a = log (x)^{ab}$.

You should know that $(a^b)^c=a^{bc}$.

3. $log_{3}x^{(1/2)4}=log_{3}x^{4/2}=log_{3}x^{2}$
i think this could help you...i'm not sure if am right but kind of having trouble in logs

You're asking lots of questions here. Let me see if I can answer them one at a time.
I have the expression $4log_{3}\sqrt[2]{x}-4log_{3}x$. In simplifying the first term I need to get rid of the square root and change the coefficient of 4 to an exponent. So if I get rid of the square root first that would be $x^{1/2}$
Correct!
so when I now change the coefficient of 4 to an exponent, shouldnt the term be $log_{3}(x^{1/2})^4$?
Yes. It is.
It seems that the actual answer is $log_{3}x^{1/2(4)}$ but according to the properties of logarithms I dont understand why.
I think that at the heart of your confusion is that you're not sure what $(x^{\frac12})^4$ means. It means:
$(x^{\frac12})\times(x^{\frac12})\times(x^{\frac12} )\times(x^{\frac12})$
which is the same as:
$x^{(\frac12+\frac12+\frac12+\frac12)}=x^{(\frac12\ times4)}$
Notice that it isn't:
$x^{(\frac12)^4}$
That would be:
$x^{(\frac12\times\frac12\times\frac12\times\frac12 )}=x^{(\frac{1}{16})}$
which is something different altogether.

The properties say that $\sqrt[x]{y}=y^{1/x}$ and $mlog n=logn^m$ so why is the simplification of the term $log_{3}x^{1/2(4)}$ and not $log_{3}(x^{1/2})^4$?
Can you see now that these are the same thing?
In other words if $mlog n=logn^m$ then why am I simply multiplying the exponent of 1/2 in $x^{1/2}$ by 4 instead of raising it to a power of 4?
See above!