Thread: logarithm

write as a single logarithm

-2log (x-2) + 1/4 log((x^2) -4) - 3log (2) + 6 log (x) =

so far I have gotten this far:

= log [ ((x^2)-4)^1/4 the whole thing over 8(x^6)(x-2)^2 ]

I am not sure what to do with the nominator I assume it will be (x-2)(x+2)^1/4 but even if that is correct I still don't know what to do next.

I am sorry about the messy writing but i don't think I have a text editor option here that would make my life easier....

Thank you

Originally Posted by zdenka

write as a single logarithm

-2log (x-2) + 1/4 log((x^2) -4) - 3log (2) + 6 log (x) =

so far I have gotten this far:

= log [ ((x^2)-4)^1/4 the whole thing over 8(x^6)(x-2)^2 ]

I am not sure what to do with the nominator I assume it will be (x-2)(x+2)^1/4 but even if that is correct I still don't know what to do next.

I am sorry about the messy writing but i don't think I have a text editor option here that would make my life easier....

Thank you

Only one change: the exponent on the 6 log(x) term is positive so it goes in the numerator:
$\displaystyle log \left ( \frac{x^6(x^2 - 4)^{1/4}}{8(x - 2)^2} \right )$

This is all you need to do with it.

-Dan

so there is nothing I can do with the Difference of squares in the nominator??


thanks

Originally Posted by zdenka

so there is nothing I can do with the Difference of squares in the nominator??


thanks

Do you mean numerator or denominator? I assume the latter.

Hello, zdenka!

Well, we can take a step further, but I don't think it's expected of us.

$\displaystyle \log\left[\frac{x^6(x^2-4)^{\frac{1}{4}}}{8(x-2)^2}\right]$

We have: .$\displaystyle \log\left[\frac{x^6(x-2)^{\frac{1}{4}}(x+2)^{\frac{1}{4}}}{8(x-2)^2}\right] \;=\;\log\left[\frac{x^6(x+2)^{\frac{1}{4}}}{8(x-2)^{\frac{7}{4}}}\right]$


But it's great that you noticed it and felt obligated to do something!

True that Soroban, and good call zdenka !

It all makes sense now..


thanks !!!!