properties of logarithms

**mgrade2000** · February 23rd 2010, 06:28 AM

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?

**Ted** · February 23rd 2010, 06:41 AM

Originally Posted by mgrade2000

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 $(a^b)^c=a^{bc}$ .

**icefirez** · February 23rd 2010, 06:50 AM

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

**Grandad** · February 23rd 2010, 07:50 AM

Hello mgrade2000

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

Originally Posted by mgrade2000

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!

Grandad

**mgrade2000** · February 24th 2010, 04:32 AM

yes it makes sense now. Thanks a lot.

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