Hello mgrade2000

You're asking lots of questions here. Let me see if I can answer them one at a time. Quote:

Originally Posted by

**mgrade2000** I have the expression $\displaystyle 4log_{3}\sqrt[2]{x}-4log_{3}x$. 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 $\displaystyle x^{1/2}$

Correct! Quote:

so when I now change the coefficient of 4 to an exponent, shouldnt the term be $\displaystyle log_{3}(x^{1/2})^4$?

Yes. It is. Quote:

It seems that the actual answer is $\displaystyle log_{3}x^{1/2(4)}$ 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 $\displaystyle (x^{\frac12})^4$ means. It means:$\displaystyle (x^{\frac12})\times(x^{\frac12})\times(x^{\frac12} )\times(x^{\frac12})$

which is the same as:$\displaystyle x^{(\frac12+\frac12+\frac12+\frac12)}=x^{(\frac12\ times4)}$

Notice that it *isn't*:$\displaystyle x^{(\frac12)^4}$

That would be:$\displaystyle x^{(\frac12\times\frac12\times\frac12\times\frac12 )}=x^{(\frac{1}{16})}$

which is something different altogether.

Quote:

The properties say that $\displaystyle \sqrt[x]{y}=y^{1/x}$ and $\displaystyle mlog n=logn^m$ so why is the simplification of the term $\displaystyle log_{3}x^{1/2(4)}$ and not $\displaystyle log_{3}(x^{1/2})^4$?

Can you see now that these are the same thing?

Quote:

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

See above!

Grandad