# logarithm help

• Jan 1st 2010, 11:57 AM
llkkjj24
logarithm 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 PM
Defunkt
Quote:

Originally Posted by llkkjj24
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

Use these two facts:

$\displaystyle log(x\cdot y) = log(x) + log(y)$

$\displaystyle log(\frac{x}{y}) = log(x) - log(y)$

And for the second, note that $\displaystyle \frac{1}{2}log(80) - \frac{1}{2}log(5) = \frac{1}{2}(log(80)-log(5))$.

Can you do it now?
• Jan 1st 2010, 12:20 PM
llkkjj24
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 PM
pickslides
Quote:

Originally Posted by llkkjj24
so was hoping for someone to work it out so i can see my mistakes

Post your work so we can see your mistakes, it will be easier to assist you this way.
• Jan 2nd 2010, 05:29 AM
llkkjj24
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 AM
Stroodle
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 AM
Stroodle
Quote:

Originally Posted by llkkjj24
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 $\displaystyle \log \frac{\log \sqrt{80}}{\log \sqrt{5}}$

You can also solve the second problem this way:

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

$\displaystyle \Rightarrow log\left ( \frac{\sqrt{80}}{\sqrt{5}}\right )$

$\displaystyle \Rightarrow log(4)$

(after rationalising and simplifying the fraction)
• Jan 2nd 2010, 08:26 AM
llkkjj24
thanks , problem solved