1.express in terms of log a, log b.

$\displaystyle \log \frac{1}{ab^4}$

2.express as a single logarithm.

$\displaystyle \frac{1}{2}\log 80 - \frac{1}{2}\log 5$

thanks

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- Jan 1st 2010, 11:57 AMllkkjj24logarithm help
1.express in terms of log a, log b.

$\displaystyle \log \frac{1}{ab^4}$

2.express as a single logarithm.

$\displaystyle \frac{1}{2}\log 80 - \frac{1}{2}\log 5$

thanks - Jan 1st 2010, 12:12 PMDefunkt
- Jan 1st 2010, 12:20 PMllkkjj24
ye, i did it before, but my answer didn't match of the book.

so was hoping for someone to work it out so i can see my mistakes - Jan 1st 2010, 12:29 PMpickslides
- Jan 2nd 2010, 05:29 AMllkkjj24
for 1 i had trouble of how to manipulate the fraction $\displaystyle \frac{1}{ab^4}$.

is this the correct form? $\displaystyle \log a - \log b ^ -4$

for 2 i did using the laws of logs.

$\displaystyle \log 80^\frac{1}{2} - \log 5^\frac{1}{2}$

then $\displaystyle \log \frac{\log \sqrt{80}}{\log \sqrt{5}}$ - Jan 2nd 2010, 06:16 AMStroodle
Using the properties posted above by Defunkt, you get:

$\displaystyle log\left ( \frac{1}{ab^4}\right )$

$\displaystyle \Rightarrow log(1)-\left[log(a)+log(b^4)\right]$

$\displaystyle \Rightarrow 0-log(a)-4log(b)$

$\displaystyle \Rightarrow -log(a)-4log(b)$

and for the second one:

$\displaystyle \frac{1}{2}log(80)-\frac{1}{2}log(5)$

$\displaystyle \Rightarrow \frac{1}{2}\left[log(80)-log(5)\right]$

$\displaystyle \Rightarrow\frac{1}{2}log(\frac{80}{5})$

$\displaystyle \Rightarrow\frac{1}{2}log(16)$

$\displaystyle \Rightarrow\log(16^\frac{1}{2})$

$\displaystyle \Rightarrow\log(4)$ - Jan 2nd 2010, 06:27 AMStroodle
- Jan 2nd 2010, 08:26 AMllkkjj24
thanks , problem solved